Monitor HDU - 6514 | 二维差分

TMD找bug找了一万年。。。一直re，跟题解对了2个小时md。
个人感想：强者就是脑子比你好使，完成同样的任务比你更快更好，然后拿出剩下的时间去享受生活(自闭时的所想，不代表个人观点)

题意：

分析：
模板背过
代码：

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
//typedef __int128 lll;
#define close() ios::sync_with_stdio(0), cin.tie(0), cout.tie(0)
#define mem(a, b) memset(a, b, sizeof(a))
const ll mod = 1e9 + 7;
const int maxn = 2e7 + 100;
const int inf = 0x3f3f3f3f;
int n, m;
int a[maxn];

int getid(int x, int y)
{
    if(x < 1 || x > n || y > m) return 0;
    return (x - 1) * m + y;
}

int main()
{
    while(cin >> n >> m)
    {
        for(int i = 0; i <= n * m; i++) a[i] = 0;
        int q;cin >> q;
        for(int i = 1; i <= q; i++)
        {
            int x1, y1, x2, y2;
            scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
            a[getid(x1, y1)]++;
            a[getid(x2 + 1, y1)]--;
            a[getid(x1, y2 + 1)]--;
            a[getid(x2 + 1, y2 + 1)]++;
        }
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= m; j++)
                a[getid(i, j)] += a[getid(i, j - 1)] + a[getid(i - 1, j)] - a[getid(i - 1, j - 1)];
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= m; j++)
                a[getid(i, j)] = a[getid(i, j)] > 0;
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= m; j++)
                a[getid(i, j)] += a[getid(i - 1, j)] + a[getid(i, j - 1)] - a[getid(i - 1, j - 1)];
        int p; cin >> p;
        while(p--)
        {
            int x1, y1, x2, y2;
            scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
            int sum = a[getid(x2, y2)] - a[getid(x2, y1 - 1)] - a[getid(x1 - 1, y2) + a[getid(x1 - 1, y1 - 1)]];
            if(sum == (x2 - x1 + 1) * (y2 - y1 + 1)) cout << "YES" << endl;
            else cout << "NO" << endl;
        }
    }
}