题目链接：点我啊╭(╯^╰)╮
题目大意：
m 1 m1 m1 次矩阵加后， m 2 m2 m2 次查询矩阵最大值
解题思路：
m 1 m1 m1 次操作可以直接差分做
问题在于怎么查询矩阵最大值
用二维线段树或者四叉树都可以做
这里给了一份比较好的模板
核心：线段树套线段树
// https://pintia.cn/problem-sets/1217641604302602240/problems/1217642214674497544
#include<bits/stdc++.h>
#define rint register int
#define deb(x) cerr<<#x<<" = "<<(x)<<'\n';
using namespace std;
typedef long long ll;
using pii = pair <int,int>;
const int maxn = 2e3 + 5;
int n, m1, m2;
ll a[maxn][maxn];
struct seg{ll t[maxn<<2];void build(int L, int l, int r, int rt){if(l == r){t[rt] = a[L][r];return;}int m = l + r >> 1;build(L, l, m, rt<<1);build(L, m+1, r, rt<<1|1);t[rt] = max(t[rt<<1], t[rt<<1|1]);}void update(int L, int R, ll x, int l, int r, int rt){t[rt] = x;if(l==L && r==R){
// lz[rt] = x;return;}int m = l + r >> 1;if(m >= R) update(L, R, x, l, m, rt<<1);else if(L > m) update(L, R, x, m+1, r, rt<<1|1);else update(L, m, x, l, m, rt<<1), \update(m+1, R, x, m+1, r, rt<<1|1);t[rt] = max(t[rt<<1], t[rt<<1|1]);}ll query(int L, int R, int l, int r, int rt, ll sum){if(l==L && r==R) return t[rt];int m = l + r >> 1;if(m >= R) return query(L, R, l, m, rt<<1, sum);else if(L > m) return query(L, R, m+1, r, rt<<1|1, sum);else return max( query(L, m, l, m, rt<<1, sum), \query(m+1, R, m+1, r, rt<<1|1, sum) );}
} t[maxn<<2], lz[maxn<<2];void merge(seg &p, seg &p1, seg &p2, int L, int R, int l, int r, int rt){if(l>R || r<L) return; p.t[rt] = max(p1.t[rt], p2.t[rt]);if(l == r) return;int m = l + r >> 1;merge(p, p1, p2, L, R, l, m, rt<<1);merge(p, p1, p2, L, R, m+1, r, rt<<1|1);
}void build(int l, int r, int rt){if(l == r){t[rt].build(l, 1, n, 1);return;}int m = l + r >> 1;build(l, m, rt<<1);build(m+1, r, rt<<1|1);merge(t[rt], t[rt<<1], t[rt<<1|1], 1, n, 1, n, 1);
}
void update(int x, int y, int xx, int yy, ll v, int l, int r, int rt){if(l==x && r==xx){t[rt].update(y, yy, v, 1, n, 1);
// lz[rt].update(y, yy, v, 1, n, 1);return;}int m = l + r >> 1;if(m >= xx) update(x, y, xx, yy, v, l, m, rt<<1);else if(x > m) update(x, y, xx, yy, v, m+1, r, rt<<1|1);else update(x, y, m, yy, v, l, m, rt<<1), \update(m+1, y, xx, yy, v, m+1, r, rt<<1|1);merge(t[rt], t[rt<<1], t[rt<<1|1], y, yy, 1, n, 1);
}ll query(int x, int y, int xx, int yy, int l, int r, int rt, ll sum){if(l==x && r==xx) return t[rt].query(y, yy, 1, n, 1, 0);
// sum += lz[rt].query(y, yy, 1, n, 1, 0);int m = l + r >> 1;if(m >= xx) return query(x, y, xx, yy, l, m, rt<<1, sum);else if(x > m) return query(x, y, xx, yy, m+1, r, rt<<1|1, sum);else return max( query(x, y, m, yy, l, m, rt<<1, sum), \query(m+1, y, xx, yy, m+1, r, rt<<1|1, sum) );
}int main() {scanf("%d%d%d", &n, &m1, &m2);int x1, x2, y1, y2, w;for(int i=1; i<=m1; i++){scanf("%d%d%d%d%d", &x1, &y1, &x2, &y2, &w);a[x1][y1] += w, a[x2+1][y2+1] += w;a[x1][y2+1] -= w, a[x2+1][y1] -= w;}for(int i=1; i<=n; i++)for(int j=1; j<=n; j++){a[i][j] += a[i][j-1] + a[i-1][j] - a[i-1][j-1];
// update(i, j, i, j, a[i][j], 1, n, 1);}build(1, n, 1);while(m2--){scanf("%d%d%d%d", &x1, &y1, &x2, &y2);printf("%lld\n", query(x1, y1, x2, y2, 1, n, 1, 0));}
}