[国家集训队]矩阵乘法

嘟嘟嘟


这题就是整体二分的板儿，只不过变成二维的。
那么就用二维树状数组好啦。
至于二维树状数组，看了代码就懂了。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("") 
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define rg register
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 505;
const int maxq = 6e4 + 5;
inline ll read()
{
 	 ll ans = 0;
  	char ch = getchar(), last = ' ';
  	while(!isdigit(ch)) last = ch, ch = getchar();
  	while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
  	if(last == '-') ans = -ans;
  	return ans;
}
inline void write(ll x)
{
	if(x < 0) x = -x, putchar('-');
	if(x >= 10) write(x / 10);
  	putchar(x % 10 + '0');
}

int n, q, cnt = 0;
struct Node
{
  	int xa, ya, xb, yb, k, id;
}t[maxn * maxn + maxq], tl[maxn * maxn + maxq], tr[maxn * maxn + maxq];
int ans[maxq];

ll c[maxn][maxn];
inline ll lowbit(const int& x) {return x & -x;}
inline void update(const int& px, const int& py, const int& d)
{
	for(int x = px; x <= n; x += lowbit(x))
		for(int y = py; y <= n; y += lowbit(y)) c[x][y] += d;	
}
inline ll query(const int& px, const int& py)
{
	ll ret = 0;
	for(int x = px; x; x -= lowbit(x))
		for(int y = py; y; y -= lowbit(y)) ret += c[x][y];
	return ret;
}
inline ll Query(const int& xa, const int& ya, const int& xb, const int& yb)
{
	return query(xb, yb) - query(xb, ya - 1) - query(xa - 1, yb) + query(xa - 1, ya - 1);	
}

inline void solve(const int& vl, const int& vr, const int& ql, const int& qr)
{
	if(ql > qr) return;
  	if(vl == vr)
    {
      	for(int i = ql; i <= qr; ++i)
      		if(t[i].id) ans[t[i].id] = vl;
      	return;
    }
    int mid = (vl + vr) >> 1, id1 = 0, id2 = 0;
    for(int i = ql; i <= qr; ++i)
    {
    	if(!t[i].id)
    	{
    		if(t[i].k <= mid) update(t[i].xa, t[i].ya, 1), tl[++id1] = t[i];
    		else tr[++id2] = t[i];
    	}
    	else
    	{
    		ll sum = Query(t[i].xa, t[i].ya, t[i].xb, t[i].yb);
    		if(sum >= t[i].k) tl[++id1] = t[i];
    		else t[i].k -= sum, tr[++id2] = t[i];
    	}
    }
    for(int i = 1; i <= id1; ++i) if(!tl[i].id && tl[i].k <= mid) update(tl[i].xa, tl[i].ya, -1);
    for(int i = 1; i <= id1; ++i) t[ql + i - 1] = tl[i];
    for(int i = 1; i <= id2; ++i) t[ql + id1 + i - 1] = tr[i];
    solve(vl, mid, ql, ql + id1 - 1);
    solve(mid + 1, vr, ql + id1, qr);
}

int main()
{
	n = read(); q = read();
	for(int i = 1; i <= n; ++i)
	    for(int j = 1; j <= n; ++j)
	    {
	    	t[++cnt].k = read();
			t[cnt].xa = i; t[cnt].ya = j; t[cnt].id = 0;
	    }
  	for(int i = 1; i <= q; ++i)
    	t[++cnt].xa = read(), t[cnt].ya = read(), t[cnt].xb = read(),
      	t[cnt].yb = read(), t[cnt].k = read(), t[cnt].id = i;
	solve(0, 1e9, 1, cnt);
	for(int i = 1; i <= q; ++i) write(ans[i]), enter;
	return 0;
}