【BZOJ 2738】矩阵乘法

题目传送门:http://www.lydsy.com/JudgeOnline/problem.php?id=2738

这个题,真的是妙啊!
整体二分看起来很好用的样子!

#include<bits/stdc++.h>
#define LL long long
#define abs(x) ((x)>0?(x):-(x))
using namespace std;

const int N = 500+9;
const int M = 250000+9;
const int Q = 60000+9;

struct Point{int val,x,y;inline bool operator < (const Point &B) const {return val < B.val;}}p[M];
struct Query{int k,x1,x2,y1,y2,id;}q[Q],buf[Q];

int n,m,vout[Q];

inline int read(){
	char c=getchar(); int ret=0,f=1;
	while (c<'0'||c>'9') {if(c=='-')f=-1;c=getchar();}
	while (c<='9'&&c>='0') ret=ret*10+c-'0',c=getchar();
	return ret*f;
}

namespace Fenwick_Tree{
	#define BIT Fenwick_Tree
	#define lowbit(x) ((x)&-(x))
	int sum[N][N];
	
	inline void modify(int x, int y, int delta) {
		if (x <= 0 || y <= 0) return;
		for (int j=y;j<=n;j+=lowbit(j)) for (int i=x;i<=n;i+=lowbit(i))
			sum[i][j] += delta;
	}
	
	inline int query(int x, int y){
		if (x <= 0 || y <= 0) return 0;
		int ret = 0;
		for (int j=y;j;j-=lowbit(j)) for (int i=x;i;i-=lowbit(i))
			ret += sum[i][j];
		return ret;
	}
};

void solve(int l, int r, int L, int R) {
	if (l == r) for (int i=L;i<=R;i++) vout[q[i].id] = p[l].val;
	else {
		int mid = l + r + 1 >> 1,ls=L-1,rs=R+1;
		for (int i=l;i<=mid-1;i++) BIT::modify(p[i].x,p[i].y,1);
		for (int i=L,tmp;i<=R;i++) {
			tmp = BIT::query(q[i].x1-1,q[i].y1-1);
			tmp += BIT::query(q[i].x2,q[i].y2);
			tmp -= BIT::query(q[i].x1-1,q[i].y2);
			tmp -= BIT::query(q[i].x2,q[i].y1-1);
			if (tmp >= q[i].k) buf[++ls] = q[i];
			else q[i].k -= tmp, buf[--rs] = q[i];
		}
		memcpy(q+L,buf+L,sizeof(buf[0])*(R-L+1));
		for (int i=l;i<=mid-1;i++) BIT::modify(p[i].x,p[i].y,-1);
		if (L <= ls) solve(l,mid-1,L,ls);
		if (rs <= R) solve(mid,r,rs,R);
	}
} 

int main(){
	n = read(); m = read();
	for (int j=1,t=1;j<=n;j++) for (int i=1;i<=n;i++,t++) 
		p[t].x = i, p[t].y = j, p[t].val = read();
	for (int i=1,a,b,c,d,e,f;i<=m;i++) 
 		q[i].y1 = read(), q[i].x1 = read(),
		q[i].y2 = read(), q[i].x2 = read(),
		q[i].k = read(), q[i].id = i;
	sort(p+1,p+1+n*n);
	solve(1,n*n,1,m);
	for (int i=1;i<=m;i++) printf("%d\n",vout[i]);
	return 0;
}