二分答案 – Tgotp-Blog

Life Forms

思路不难想，求出sa以后二分随便搞搞，和上一题比较像。细节卡死人。。。

注意是大于一半。。。一开始以为是至少一半疯狂wa。。。

然后现在还把用来对拍的程序卡掉了。。。

c++代码如下：

/* code by Tgotp
    poj 3294  */
#include<algorithm>
#include<cstdio>
#include<cstring>
#define rep(i,x,y) for(register int i = x ;i <= y; ++ i)
#define repd(i,x,y) for(register int i = x ; i >= y; -- i)
using namespace std;

const int N = 1e5 + 500;
char s[1011],S[N];
int len,x[N],y[N << 1],sa[N],buc[N],height[N],rank[N];
int n,pos[111],belong[N];

inline void clear()
{
	rep(i,0,len - 1) S[i] = 0;
	memset(buc,0,sizeof buc);
	memset(x,0,sizeof x);
	len = 0;
}

inline void swap(int&x,int&y) { int t = x; x = y; y = t; }

inline void SA()
{
	int m = 455;
	rep(i,0,len - 1) ++buc[x[i]];
	rep(i,1,m) buc[i] += buc[i - 1];
	repd(i,len - 1,0) sa[--buc[x[i]]] = i;
	for(register int k = 1 ;k <= len; k <<= 1)
	{
		int p = 0;
		repd(i,len - 1,len - k) y[p++] = i;
		rep(i,0,len - 1) if(sa[i] >= k) y[p++] = sa[i] - k;
		
		rep(i,0,m) buc[i] = 0;
		rep(i,0,len - 1) ++buc[x[i]];
		rep(i,1,m) buc[i] += buc[i - 1];
		repd(i,len - 1,0) sa[--buc[x[y[i]]]] = y[i];
		
		rep(i,0,len - 1) swap(x[i],y[i]);
		p = 1; x[sa[0]] = 0;
		rep(i,1,len - 1)
			if(y[sa[i]] == y[sa[i - 1]] && y[sa[i] + k] == y[sa[i - 1] + k])
				x[sa[i]] = p - 1;
			else x[sa[i]] = p++;
		
		if(p == len) break;
		m = p;
	}
	
	int p = 0;
	rep(i,0,len - 1) rank[sa[i]] = i;
	rep(i,0,len - 1)
	{
		if(p) --p;
		if(rank[i] == 0) continue;
		int t = sa[rank[i] - 1];
		while(t + p < len && i + p < len && S[t + p] == S[i + p] ) ++p;
		height[rank[i]] = p;
	}
}

inline bool check(int x)
{
	rep(i,1,len - 1)
	{
		int l,r;
		l = i; while(height[l] < x && l < len) ++l;
		r = l; while(height[r] >= x && r < len) ++r;
		
		if(l == len) break;
		
		rep(k,1,n) pos[k] = -1;
		rep(k,l - 1,r - 1)
			pos[belong[sa[k]]] = sa[k];
		
		int num = 0;
		rep(k,1,n) if(~pos[k]) ++num;
		if(num > n / 2) return true;
		
		i = r;
	}
	return false;
}

inline void solve()
{
	clear();
	
	rep(i,1,n)
	{
		scanf("%s",s);
		int Len = strlen(s);
		rep(j,0,Len - 1) x[j + len] = s[j] + 100,S[j + len] = s[j],belong[j + len] = i;
		x[len + Len] = i;S[len + Len] = i;
		belong[len + Len] = 0;
		len += Len + 1;
	}
	
	SA();
	
	int l = 0,r = len - 1,mid,ans;
	
	while(l <= r )
	{
		if(check(mid = l + r >> 1)) l = mid + 1,ans = mid;
		else r = mid - 1;
	}
	
	if(ans == 0) puts("?");
	else 
	{
		rep(i,1,len - 1)
		{
			int l,r;
			l = i; while(height[l] < ans && l < len) ++l;
			r = l; while(height[r] >= ans && r < len) ++r;
			
			if(l == len) break;
			
			rep(k,1,n) pos[k] = -1;
			rep(k,l - 1,r - 1)
				pos[belong[sa[k]]] = sa[k];
			
			int num = 0;
			rep(k,1,n) if(~pos[k]) ++num;
			if(num > n / 2) 
			{
				rep(k,0,ans - 1) printf("%c",S[sa[l - 1] + k]);
				puts("");
			}
			
			i = r;
		}
	}
	puts("");
}

int main()
{
//	freopen("1.in","r",stdin);
//	freopen("my.out","w",stdout);
	while(scanf("%d",&n),n != 0) solve();
	return 0;
}

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/* code by Tgotp

poj 3294 */

#include<algorithm>

#include<cstdio>

#include<cstring>

#define rep(i,x,y) for(register int i = x ;i <= y; ++ i)

#define repd(i,x,y) for(register int i = x ; i >= y; -- i)

using namespace std;

const int N = 1e5 + 500;

char s[1011],S[N];

int len,x[N],y[N << 1],sa[N],buc[N],height[N],rank[N];

int n,pos[111],belong[N];

inline void clear()

{

rep(i,0,len - 1) S[i] = 0;

memset(buc,0,sizeof buc);

memset(x,0,sizeof x);

len = 0;

}

inline void swap(int&x,int&y) { int t = x; x = y; y = t; }

inline void SA()

{

int m = 455;

rep(i,0,len - 1) ++buc[x[i]];

rep(i,1,m) buc[i] += buc[i - 1];

repd(i,len - 1,0) sa[--buc[x[i]]] = i;

for(register int k = 1 ;k <= len; k <<= 1)

{

int p = 0;

repd(i,len - 1,len - k) y[p++] = i;

rep(i,0,len - 1) if(sa[i] >= k) y[p++] = sa[i] - k;

rep(i,0,m) buc[i] = 0;

rep(i,0,len - 1) ++buc[x[i]];

rep(i,1,m) buc[i] += buc[i - 1];

repd(i,len - 1,0) sa[--buc[x[y[i]]]] = y[i];

rep(i,0,len - 1) swap(x[i],y[i]);

p = 1; x[sa[0]] = 0;

rep(i,1,len - 1)

if(y[sa[i]] == y[sa[i - 1]] && y[sa[i] + k] == y[sa[i - 1] + k])

x[sa[i]] = p - 1;

else x[sa[i]] = p++;

if(p == len) break;

m = p;

}

int p = 0;

rep(i,0,len - 1) rank[sa[i]] = i;

rep(i,0,len - 1)

{

if(p) --p;

if(rank[i] == 0) continue;

int t = sa[rank[i] - 1];

while(t + p < len && i + p < len && S[t + p] == S[i + p] ) ++p;

height[rank[i]] = p;

}

inline bool check(int x)

{

rep(i,1,len - 1)

{

int l,r;

l = i; while(height[l] < x && l < len) ++l;

r = l; while(height[r] >= x && r < len) ++r;

if(l == len) break;

rep(k,1,n) pos[k] = -1;

rep(k,l - 1,r - 1)

pos[belong[sa[k]]] = sa[k];

int num = 0;

rep(k,1,n) if(~pos[k]) ++num;

if(num > n / 2) return true;

i = r;

}

return false;

}

inline void solve()

{

clear();

rep(i,1,n)

{

scanf("%s",s);

int Len = strlen(s);

rep(j,0,Len - 1) x[j + len] = s[j] + 100,S[j + len] = s[j],belong[j + len] = i;

x[len + Len] = i;S[len + Len] = i;

belong[len + Len] = 0;

len += Len + 1;

}

SA();

int l = 0,r = len - 1,mid,ans;

while(l <= r )

{

if(check(mid = l + r >> 1)) l = mid + 1,ans = mid;

else r = mid - 1;

}

if(ans == 0) puts("?");

else

{

rep(i,1,len - 1)

{

int l,r;

l = i; while(height[l] < ans && l < len) ++l;

r = l; while(height[r] >= ans && r < len) ++r;

if(l == len) break;

rep(k,1,n) pos[k] = -1;

rep(k,l - 1,r - 1)

pos[belong[sa[k]]] = sa[k];

int num = 0;

rep(k,1,n) if(~pos[k]) ++num;

if(num > n / 2)

{

rep(k,0,ans - 1) printf("%c",S[sa[l - 1] + k]);

puts("");

}

i = r;

}

puts("");

}

int main()

{

// freopen("1.in","r",stdin);

// freopen("my.out","w",stdout);

while(scanf("%d",&n),n != 0) solve();

return 0;

}

数据生成代码：
[cra[......]

2017年12月22日 241 0 /

poj

Tag: sa 二分答案

PHRASES - Relevant Phrases of Annihilation

有点难度的sa，把所有的字符串串在一起，然后分段保证段上有长为x的两个子串

首先求出sa，然后再求出height 数组，然后二分答案，对于每段就枚举height差值不超过二分值得段。

然后记录在每一段出现最早与最晚得值（判断有无两个串），然后判断。

搞定。

c++代码如下：[......]

2017年12月21日 254 0 /

SPOJ

Tag: sa 二分答案

P3939 数颜色

因为交换只会交换相邻两数，所以交换很好写。

用vector存一下每个颜色出现的位置，然后对于查找的颜色在vector上二分。

c++代码如下：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<queue>
#include<stack>
#include<vector>
#include<algorithm>
#define repd(i,x,y) for(register int i = x ;i >= y; --i)
#define rep(i,x,y) for(register int i = x ;i <= y;++i)
using namespace std;
typedef long long ll;
template<typename T>inline void read(T&x)
{
    x = 0;char c;int sign = 1;
    do { c = getchar(); if(c == '-') sign = -1; }while(c < '0' || c > '9');
    do { x = x * 10 + c - '0'; c = getchar(); }while(c <= '9' && c >= '0');
    x *= sign;
}

const int N = 3e5 + 500;
int n,m,a[N],id[N];
vector<int>color[N];

int main()
{
	read(n);read(m);
	
	rep(i,1,3e5)color[i].push_back(0);
	rep(i,1,n)
	{
		read(a[i]);
		color[a[i]].push_back(i);
		id[i] = color[a[i]].size() - 1;
	}
	rep(i,1,3e5)color[i].push_back(n + 1);
	
	int op,l,r,x;
	rep(i,1,m)
	{
		read(op);
		if(op == 1)
		{
			read(l);read(r);read(x);
			int L = 0,R = ((int)color[x].size()) - 1,mid,z1 = 0,z2 = -1;
			while(L <= R)
			{
				if(color[x][mid = (L + R) >> 1] < l) z1 = mid,L = mid + 1;
				else R = mid - 1;
			}
			L = 0,R = ((int)color[x].size()) - 1;
			while(L <= R)
			{
				if(color[x][mid = (L + R) >> 1] > r) z2 = mid,R = mid - 1;
				else L = mid + 1;
			}
			printf("%d\n",z2 - z1 - 1 > 0 ? z2 - z1 - 1 : 0 );
		}
		else 
		{
			read(x);
			if(a[x] == a[x + 1]) continue;
			color[ a[ x ] ][ id[ x ] ]++;
			color[a[x + 1]][id[x + 1]]--;
			swap(a[x],a[x+1]);swap(id[x],id[x + 1]);
		}
	}
	return 0;
}

#include<iostream>

#include<cstdio>

#include<cstring>

#include<string>

#include<queue>

#include<stack>

#include<vector>

#include<algorithm>

#define repd(i,x,y) for(register int i = x ;i >= y; --i)

#define rep(i,x,y) for(register int i = x ;i <= y;++i)

using namespace std;

typedef long long ll;

template<typename T>inline void read(T&x)

{

x = 0;char c;int sign = 1;

do { c = getchar(); if(c == '-') sign = -1; }while(c < '0' || c > '9');

do { x = x * 10 + c - '0'; c = getchar(); }while(c <= '9' && c >= '0');

x *= sign;

}

const int N = 3e5 + 500;

int n,m,a[N],id[N];

vector<int>color[N];

int main()

{

read(n);read(m);

rep(i,1,3e5)color[i].push_back(0);

rep(i,1,n)

{

read(a[i]);

color[a[i]].push_back(i);

id[i] = color[a[i]].size() - 1;

}

rep(i,1,3e5)color[i].push_back(n + 1);

int op,l,r,x;

rep(i,1,m)

{

read(op);

if(op == 1)

{

read(l);read(r);read(x);

int L = 0,R = ((int)color[x].size()) - 1,mid,z1 = 0,z2 = -1;

while(L <= R)

{

if(color[x][mid = (L + R) >> 1] < l) z1 = mid,L = mid + 1;

else R = mid - 1;

}

L = 0,R = ((int)color[x].size()) - 1;

while(L <= R)

{

if(color[x][mid = (L + R) >> 1] > r) z2 = mid,R = mid - 1;

else L = mid + 1;

}

printf("%d\n",z2 - z1 - 1 > 0 ? z2 - z1 - 1 : 0 );

}

else

{

read(x);

if(a[x] == a[x + 1]) continue;

color[ a[ x ] ][ id[ x ] ]++;

color[a[x + 1]][id[x + 1]]--;

swap(a[x],a[x+1]);swap(id[x],id[x + 1]);

}

return 0;

}

2017年11月2日 406 0 /

Luogu

Tag: 二分答案

3110: [Zjoi2013]K大数查询

整体二分，用线段树维护区间大于mid 的值

c++代码如下：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define rep(i,x,y) for(register int i = x ;i <= (y);++i)
#define repd(i,x,y) for(register int i = x ;i >= (y); -- i)
#define abs(x) (x > 0 ? x : -(x))
using namespace std;
typedef long long ll;
template<class T> inline void chkmin(T&a,T b) { return a > b ? a = b,1 : 0; }
template<class T> inline void chkmax(T&a,T b) { return a < b ? a = b,1 : 0; }
template<typename T>inline void read(T&x){
    x = 0;char c;int sign = 1;
    do { c = getchar(); if(c == '-') sign = -1; }while(c < '0' || c > '9');
    do { x = x * 10 + c - '0'; c = getchar(); }while(c <= '9' &&  c >= '0');
    x *= sign;
}

const int M = 5e4 + 50;

int ans[M],n,m;
struct QUE
{
    int l,r,id,op,v;
    bool k;
}q[M];

struct Segment_tree
{
    int ql,qr;
	ll add[M<<2],sum[M<<2];
    bool clr[M<<2];
    
    #define lc id<<1
    #define rc id<<1|1
    inline void init() { sum[1] = add[1] = 0;clr[1] = 1; }
    inline void update(int id) { sum[id] = sum[lc] + sum[rc]; }
    inline void putdown(int id,int l,int r)
    {
        if(clr[id])
        {
            sum[lc] = sum[rc] = add[lc] = add[rc] = 0;
            clr[lc] = clr[rc] = 1;clr[id] = 0;
        }
        int mid = (l + r) >> 1;
        if(add[id])
        {
            add[lc] += add[id]; add[rc] += add[id];
            sum[lc] += (mid - l + 1)*add[id];sum[rc] += (r - mid)*add[id];
            add[id] = 0;
        }
    }
    void Add(int id,int l,int r)
    {
        if(ql <= l && qr >= r) { add[id]++,sum[id] += r - l + 1 ; return; }
        int mid = (l + r) >> 1;
		putdown(id,l,r);
        if(ql <= mid) Add(lc,l,mid); if(qr > mid) Add(rc,mid + 1,r);
        update(id);
    }
    ll query(int id,int l,int r)
    {
        if(ql <= l && qr >= r)  return sum[id];
        ll res = 0;int mid = (l + r) >> 1;
		putdown(id,l,r);
        if(ql <= mid) res += query(lc,l,mid); if(qr > mid) res += query(rc,mid + 1,r);
        return res;
    }
}seg;

inline bool cmp(QUE a,QUE b){ return a.k == b.k ? a.id < b.id : a.k < b.k; }

void solve(int L,int R,int l,int r)
{
    if(l > r) return;
    if(L == R) {
        rep(i,l,r) if(q[i].op == 2) ans[q[i].id] = L;
        return ;
    }
    seg.init();
    int mid = (L + R ) >> 1,t = l - 1;
	ll s;
    rep(i,l,r)
    {
        if(q[i].op == 1){
            if(q[i].v > mid){
                seg.ql = q[i].l;seg.qr = q[i].r;
                seg.Add(1,1,n); q[i].k = 1;
            }else t++,q[i].k = 0;
        }
        else 
        {
            seg.ql = q[i].l;seg.qr = q[i].r;
            s = seg.query(1,1,n);
            if(q[i].v <= s ) q[i].k = 1 ;
            else t++,q[i].k = 0,q[i].v -= s;
        }
    }
    sort(q + l, q + r + 1,cmp);
    solve(L,mid,l,t) ; solve(mid + 1,R,t + 1,r);
}

int main()
{
    memset(ans,-1,sizeof ans );
    read(n);read(m);
    rep(i,1,m) {
        read(q[i].op) ; read(q[i].l);read(q[i].r);read(q[i].v);
        q[i].id = i;
    }
    solve(1,n,1,m);
    rep(i,1,m) if(ans[i] != -1)
            printf("%d\n",ans[i]);
    return 0;
}

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#include<iostream>

#include<cstdio>

#include<cstring>

#include<algorithm>

#define rep(i,x,y) for(register int i = x ;i <= (y);++i)

#define repd(i,x,y) for(register int i = x ;i >= (y); -- i)

#define abs(x) (x > 0 ? x : -(x))

using namespace std;

typedef long long ll;

template<class T> inline void chkmin(T&a,T b) { return a > b ? a = b,1 : 0; }

template<class T> inline void chkmax(T&a,T b) { return a < b ? a = b,1 : 0; }

template<typename T>inline void read(T&x){

x = 0;char c;int sign = 1;

do { c = getchar(); if(c == '-') sign = -1; }while(c < '0' || c > '9');

do { x = x * 10 + c - '0'; c = getchar(); }while(c <= '9' && c >= '0');

x *= sign;

}

const int M = 5e4 + 50;

int ans[M],n,m;

struct QUE

{

int l,r,id,op,v;

bool k;

}q[M];

struct Segment_tree

{

int ql,qr;

ll add[M<<2],sum[M<<2];

bool clr[M<<2];

#define lc id<<1

#define rc id<<1|1

inline void init() { sum[1] = add[1] = 0;clr[1] = 1; }

inline void update(int id) { sum[id] = sum[lc] + sum[rc]; }

inline void putdown(int id,int l,int r)

{

if(clr[id])

{

sum[lc] = sum[rc] = add[lc] = add[rc] = 0;

clr[lc] = clr[rc] = 1;clr[id] = 0;

}

int mid = (l + r) >> 1;

if(add[id])

{

add[lc] += add[id]; add[rc] += add[id];

sum[lc] += (mid - l + 1)*add[id];sum[rc] += (r - mid)*add[id];

add[id] = 0;

}

void Add(int id,int l,int r)

{

if(ql <= l && qr >= r) { add[id]++,sum[id] += r - l + 1 ; return; }

int mid = (l + r) >> 1;

putdown(id,l,r);

if(ql <= mid) Add(lc,l,mid); if(qr > mid) Add(rc,mid + 1,r);

update(id);

}

ll query(int id,int l,int r)

{

if(ql <= l && qr >= r) return sum[id];

ll res = 0;int mid = (l + r) >> 1;

putdown(id,l,r);

if(ql <= mid) res += query(lc,l,mid); if(qr > mid) res += query(rc,mid + 1,r);

return res;

}

}seg;

inline bool cmp(QUE a,QUE b){ return a.k == b.k ? a.id < b.id : a.k < b.k; }

void solve(int L,int R,int l,int r)

{

if(l > r) return;

if(L == R) {

rep(i,l,r) if(q[i].op == 2) ans[q[i].id] = L;

return ;

}

seg.init();

int mid = (L + R ) >> 1,t = l - 1;

ll s;

rep(i,l,r)

{

if(q[i].op == 1){

if(q[i].v > mid){

seg.ql = q[i].l;seg.qr = q[i].r;

seg.Add(1,1,n); q[i].k = 1;

}else t++,q[i].k = 0;

}

else

{

seg.ql = q[i].l;seg.qr = q[i].r;

s = seg.query(1,1,n);

if(q[i].v <= s ) q[i].k = 1 ;

else t++,q[i].k = 0,q[i].v -= s;

}

sort(q + l, q + r + 1,cmp);

solve(L,mid,l,t) ; solve(mid + 1,R,t + 1,r);

}

int main()

{

memset(ans,-1,sizeof ans );

read(n);read(m);

rep(i,1,m) {

read(q[i].op) ; read(q[i].l);read(q[i].r);read(q[i].v);

q[i].id = i;

}

solve(1,n,1,m);

rep(i,1,m) if(ans[i] != -1)

printf("%d\n",ans[i]);

return 0;

}

2017年10月12日 246 0 /

BZOJ

Tag: 二分答案整体二分线段树

P2538 [SCOI2008]城堡

爆搜 + 贪心 + 二分 + floyd；

记住：大胆假设，不用证明

先存图，观察数据范围n <= 50 ，直接用floyd求出最短路。

继续二分最大长度 c ，对于每个已有城堡的城市，直接去标记其能到达的城市，然后对于不能到达的，

我们将其距离不超过枚举的c的点期望[......]

2017年9月27日 591 0 /

Luogu

Tag: floyd 二分答案期望

1014: [JSOI2008]火星人prefix

一脸萎靡。。。

这题做的疯。

数据真心有毒。

数字和字母之前不等距 exm？

二分 + hash + splay

我只能说是个p的神题。

bz的常数卡的我疯狂。。。

就是一般的splay入门题，看看代码就好了。

c++代码如下：
[crayon-5a[......]

2017年9月22日 435 0 /

BZOJ

Tag: hash Splay 二分答案

P1024 一元三次方程求解

回归刷水题，二分的水题，直接见代码，可以当成二分答案的入门题了。

c++代码如下：

#include<iostream>
#include<cstdio>
#define rep(i,x,y) for(register int i = x;i <= y;++i)
using namespace std;
double a,b,c,d;
inline double f(double x){ return a*x*x*x+b*x*x+c*x+d; }
int main()
{
	scanf("%lf%lf%lf%lf",&a,&b,&c,&d);
	rep(i,-100,100)
	{
		double l = i,r = i + 1,mid;
		if(f(l) == 0) { printf("%.2lf ",l); continue; }
		if(f(l) * f(r) < 0)
		{
			while(r - l > 0.00001)
			{
				if(f(l) * f(mid = (l+r)/2) < 0) r = mid;
				else l = mid;
			}
			printf("%.2lf ",l);
		}
	}
	return 0;
}

#include<iostream>

#include<cstdio>

#define rep(i,x,y) for(register int i = x;i <= y;++i)

using namespace std;

double a,b,c,d;

inline double f(double x){ return a*x*x*x+b*x*x+c*x+d; }

int main()

{

scanf("%lf%lf%lf%lf",&a,&b,&c,&d);

rep(i,-100,100)

{

double l = i,r = i + 1,mid;

if(f(l) == 0) { printf("%.2lf ",l); continue; }

if(f(l) * f(r) < 0)

{

while(r - l > 0.00001)

{

if(f(l) * f(mid = (l+r)/2) < 0) r = mid;

else l = mid;

}

printf("%.2lf ",l);

}

return 0;

}

2017年8月27日 748 0 /

Luogu

Tag: 二分答案

BZOJ1067: [SCOI2007]降雨量

时隔半个月写的第一题，还是改的原来的。。。

这道题算法很显而易见，重点在于细节处理。

建代码的处理咯

c++代码如下：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
#define rep(i,x,y) for(register int i = x;i <= y;++i)
using namespace std;
inline void read(int&x)
{
	x = 0; char c;int sign = 1;
	do { c = getchar();if(c == '-') sign = -1; }while(c < '0' || c > '9');
	do { x = x*10 + c - '0';c = getchar(); }while(c <= '9' && c >= '0');
	x *= sign;
}

const int N = 5e4 + 50;
int n,m,x,id[N],val[N];

struct node
{
	int left,right,val;
}tree[N*4];

void build(int id,int l,int r)
{
	tree[id].left = l;tree[id].right = r;
	if(l == r)  { tree[id].val = val[l]; return ; }
	int mid = (l + r) >>1;
	build(id<<1,l,mid);build(id<<1|1,mid+1,r);
	tree[id].val = max(tree[id<<1].val,tree[id<<1|1].val);
}

int query(int id,int l,int r)
{
	if(l == tree[id].left && tree[id].right == r) return tree[id].val;
	int mid = (tree[id].left + tree[id].right) >>1;
	if(l > mid) return query(id<<1|1,l,r);
	if(r <= mid) return query(id<<1,l,r);
	return max(query(id<<1,l,mid),query(id<<1|1,mid+1,r));
}
inline int find(int x)
{
	int l = 1,r = n,mid,num;
	while(l <= r)
	{
		if(id[mid = (l+r)>>1] == x) return mid;
		else if(id[mid] < x) l = mid + 1;
		else r = mid - 1;
	}
	return l;
}
int main()
{
	read(n);
	int x,y;
	rep(i,1,n)
	{
		read(x);read(y);
		id[i] = x;val[i] = y;
	}
	build(1,1,n);
	read(m);
	while(m--)
	{
		bool e = 0; 
		read(y); read(x);
		if(y > x) { puts("false");continue; } 
		if(x == y) { puts("true");continue; } 
		int Y = find(y), X = find(x);int Z = Y;
		if(id[X] != x){ 
			if(Y + 1 <= X - 1 && id[Y] == y) puts(query(1,Y+1,X-1) >= val[Y] ? "false":"maybe");
			else puts("maybe");
			continue; 
		}
		if(y == id[Y]){
			if(val[Y] < val[X]) { puts("false");continue; }
			++Y;
		}else e = 1;
		if(X == Y){ puts(x - y == X - Z ?"true":"maybe"); continue; }
		int ans = query(1,Y,X-1);
		puts(val[X] <= ans?"false":x - y == X - Z && !e ?"true":"maybe");
	}
	return 0;
}

#include<iostream>

#include<cstdio>

#include<cstring>

#include<queue>

#include<algorithm>

#define rep(i,x,y) for(register int i = x;i <= y;++i)

using namespace std;

inline void read(int&x)

{

x = 0; char c;int sign = 1;

do { c = getchar();if(c == '-') sign = -1; }while(c < '0' || c > '9');

do { x = x*10 + c - '0';c = getchar(); }while(c <= '9' && c >= '0');

x *= sign;

}

const int N = 5e4 + 50;

int n,m,x,id[N],val[N];

struct node

{

int left,right,val;

}tree[N*4];

void build(int id,int l,int r)

{

tree[id].left = l;tree[id].right = r;

if(l == r) { tree[id].val = val[l]; return ; }

int mid = (l + r) >>1;

build(id<<1,l,mid);build(id<<1|1,mid+1,r);

tree[id].val = max(tree[id<<1].val,tree[id<<1|1].val);

}

int query(int id,int l,int r)

{

if(l == tree[id].left && tree[id].right == r) return tree[id].val;

int mid = (tree[id].left + tree[id].right) >>1;

if(l > mid) return query(id<<1|1,l,r);

if(r <= mid) return query(id<<1,l,r);

return max(query(id<<1,l,mid),query(id<<1|1,mid+1,r));

}

inline int find(int x)

{

int l = 1,r = n,mid,num;

while(l <= r)

{

if(id[mid = (l+r)>>1] == x) return mid;

else if(id[mid] < x) l = mid + 1;

else r = mid - 1;

}

return l;

}

int main()

{

read(n);

int x,y;

rep(i,1,n)

{

read(x);read(y);

id[i] = x;val[i] = y;

}

build(1,1,n);

read(m);

while(m--)

{

bool e = 0;

read(y); read(x);

if(y > x) { puts("false");continue; }

if(x == y) { puts("true");continue; }

int Y = find(y), X = find(x);int Z = Y;

if(id[X] != x){

if(Y + 1 <= X - 1 && id[Y] == y) puts(query(1,Y+1,X-1) >= val[Y] ? "false":"maybe");

else puts("maybe");

continue;

}

if(y == id[Y]){

if(val[Y] < val[X]) { puts("false");continue; }

++Y;

}else e = 1;

if(X == Y){ puts(x - y == X - Z ?"true":"maybe"); continue; }

int ans = query(1,Y,X-1);

puts(val[X] <= ans?"false":x - y == X - Z && !e ?"true":"maybe");

}

return 0;

}

2017年8月20日 505 0 /

BZOJ

Tag: 二分答案线段树

疫情控制 NOIP2012 T6

二分答案 + 贪心；

你会发现，其实对于每个点，我们只要能往上（不是根节点）一定是更优的，如果是根节点，我们先看看这个点通不通，不通的话这个点判通，返回，不然的话进入根节点找最小不通的点。

最后一个点恶心的菊花图，所以对于这道题的极限数据，你倍增不倍增都没卵用，我也懒得帖我倍增+桶排的代[......]

2017年7月3日 522 0 /

Luogu NOIP/NOI

Tag: 二分答案倍增贪心法

聪明的质监员 NOIP2011

二分答案，sum[i]表示前i个中符合>=x的个数，num[i]表示前i个中这些符合的v之和；剩下的直接二分搞搞就好了。原谅sum与num的英语意义乱用2333.

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
using namespace std;
typedef long long ll;
inline void read(int&x)
{
	x = 0; char c ; int sign = 1;
	do{ c = getchar(); if(c == '-') sign = -1; }while(c < '0' || c > '9');
	do{ x = x * 10 + c - '0';c = getchar(); }while(c <= '9' && c >= '0');
	x *= sign;
}
const int N = 2e5+5;
int n ,m ,mid,sum[N];
ll num[N],ans = 1e16,y;
struct node{
	int w,v;
}a[N];
struct str{
	int l,r;
}p[N];
ll kk(ll x){ return x >= 0? x: -x; }
ll min(ll x,ll y){ return x > y? y : x; }
bool check(int x)
{
	ll z = 0;
	memset(num,0,sizeof(num)); memset(sum,0,sizeof(sum));
	for(int i = 1 ;i <= n; i++) { num[i] = num[i-1];sum[i] = sum[i-1]; if(a[i].w >= x) sum[i]++,num[i]+= a[i].v; }
	for(int i = 1 ;i <= m; i++)
		z += (sum[p[i].r] - sum[p[i].l - 1])*(num[p[i].r] - num[p[i].l - 1]);
	ans = min(ans,kk(z - y));
	if(z < y) return false;
	else return true;
}
int main()
{
	read(n);read(m);scanf("%lld",&y);
	int l = 0,r = 0;
	for(int i = 1 ;i <= n;i++) { read(a[i].w);read(a[i].v); r = max(r,a[i].w);  }
	for(int i = 1 ;i <= m;i++) { read(p[i].l);read(p[i].r); }
	while(l <= r)
	{
		if(check(mid = (l+r) >>1)) l = mid + 1;
		else r = mid - 1 ;
	}
	cout<<ans;
	return 0;
}

#include<iostream>

#include<cstring>

#include<algorithm>

#include<cstdio>

using namespace std;

typedef long long ll;

inline void read(int&x)

{

x = 0; char c ; int sign = 1;

do{ c = getchar(); if(c == '-') sign = -1; }while(c < '0' || c > '9');

do{ x = x * 10 + c - '0';c = getchar(); }while(c <= '9' && c >= '0');

x *= sign;

}

const int N = 2e5+5;

int n ,m ,mid,sum[N];

ll num[N],ans = 1e16,y;

struct node{

int w,v;

}a[N];

struct str{

int l,r;

}p[N];

ll kk(ll x){ return x >= 0? x: -x; }

ll min(ll x,ll y){ return x > y? y : x; }

bool check(int x)

{

ll z = 0;

memset(num,0,sizeof(num)); memset(sum,0,sizeof(sum));

for(int i = 1 ;i <= n; i++) { num[i] = num[i-1];sum[i] = sum[i-1]; if(a[i].w >= x) sum[i]++,num[i]+= a[i].v; }

for(int i = 1 ;i <= m; i++)

z += (sum[p[i].r] - sum[p[i].l - 1])*(num[p[i].r] - num[p[i].l - 1]);

ans = min(ans,kk(z - y));

if(z < y) return false;

else return true;

}

int main()

{

read(n);read(m);scanf("%lld",&y);

int l = 0,r = 0;

for(int i = 1 ;i <= n;i++) { read(a[i].w);read(a[i].v); r = max(r,a[i].w); }

for(int i = 1 ;i <= m;i++) { read(p[i].l);read(p[i].r); }

while(l <= r)

{

if(check(mid = (l+r) >>1)) l = mid + 1;

else r = mid - 1 ;

}

cout<<ans;

return 0;

}

2017年7月3日 508 0 /

Luogu NOIP/NOI

Tag: 二分答案前缀和