2018徐州 Gym - 102028J Carpets Removal 概率事件+二維差分

題目鏈接：https://vjudge.net/problem/Gym-102028J

題意：去掉兩個矩形，讓剩下的矩形覆蓋的點最少

題解：使剩下的點少，那麼我們就是去掉的兩個矩形覆蓋的儘可能多，也就是兩個矩形各自佔據的格子和只有兩個矩形佔據的格子，像這種隨機數據，去掉的最優的情況很容易就在各自佔用很多的情況，或者兩者同時覆蓋很多的情況，那麼我們就按照幾種情況排序只處理前幾百個矩形就行。

#include<bits/stdc++.h>
using namespace std;
const int N = 3e5 + 100;
struct node {
    int x1, x2, y1, y2;
    int val1, val2;
}a[N];
int n, m;
int vis[1510][1510];
int sum1[1510][1510], sum2[1510][1510];
int cmp1(const node &x, const node &y) {
    return x.val1 > y.val1;
}
int cmp2(const node &x, const node &y) {
    return x.val2 > y.val2;
}
int cmp3(const node &x, const node &y) {
    return x.val1 + x.val2 > y.val1 + y.val2;
}
int main() {
    int T;
    int ans, minn, res;
    int cnt;
    int x1, x2, y1, y2;
    scanf("%d", &T);
    while(T--) {
        scanf("%d %d", &n, &m);
        for(int i = 0; i <= m; i++)
            for(int j = 0; j <= m; j++)
                vis[i][j] = sum1[i][j] = sum2[i][j] = 0;
        for(int i = 1; i <= n; i++) {
            scanf("%d %d %d %d", &a[i].x1, &a[i].x2, &a[i].y1, &a[i].y2);
            vis[a[i].x1][a[i].y1]++;
            vis[a[i].x1][a[i].y2 + 1]--;
            vis[a[i].x2 + 1][a[i].y1]--;
            vis[a[i].x2 + 1][a[i].y2 + 1]++;
        }
        ans = 0;

        for(int i = 1; i <= m; i++)
            for(int j = 1; j <= m; j++) {
                vis[i][j] += vis[i - 1][j] + vis[i][j - 1] - vis[i - 1][j - 1];
                sum1[i][j] += sum1[i - 1][j] + sum1[i][j - 1] - sum1[i - 1][j - 1];
                sum2[i][j] += sum2[i - 1][j] + sum2[i][j - 1] - sum2[i - 1][j - 1];
                if(vis[i][j] == 1) sum1[i][j]++;
                if(vis[i][j] == 2) sum2[i][j]++;
                if(vis[i][j]) ans++;
            }
        minn = ans;
        for(int i = 1; i <= n; i++) {
            a[i].val1 = sum1[a[i].x2][a[i].y2] - sum1[a[i].x1 - 1][a[i].y2] - sum1[a[i].x2][a[i].y1 - 1] + sum1[a[i].x1 - 1][a[i].y1 - 1];
            a[i].val2 = sum2[a[i].x2][a[i].y2] - sum2[a[i].x1 - 1][a[i].y2] - sum2[a[i].x2][a[i].y1 - 1] + sum2[a[i].x1 - 1][a[i].y1 - 1];
        }
        cnt = min(n, 100);
        sort(a + 1, a + 1 + n, cmp1);
        for(int i = 1; i <= cnt; i++)
            for(int j = i + 1; j <= cnt; j++) {
            x1 = max(a[i].x1, a[j].x1);
            x2 = min(a[i].x2, a[j].x2);
            y1 = max(a[i].y1, a[j].y1);
            y2 = min(a[i].y2, a[j].y2);
            res = a[i].val1 + a[j].val1;
            if(x1 <= x2 && y1 <= y2) {
                res += sum2[x2][y2] - sum2[x1 - 1][y2] - sum2[x2][y1 - 1] + sum2[x1 - 1][y1 - 1];
            }
            minn = min(minn, ans - res);
        }
        sort(a + 1, a + 1 + n, cmp2);
        for(int i = 1; i <= cnt; i++)
            for(int j = i + 1; j <= cnt; j++) {
            x1 = max(a[i].x1, a[j].x1);
            x2 = min(a[i].x2, a[j].x2);
            y1 = max(a[i].y1, a[j].y1);
            y2 = min(a[i].y2, a[j].y2);
            res = a[i].val1 + a[j].val1;
            if(x1 <= x2 && y1 <= y2) {
                res += sum2[x2][y2] - sum2[x1 - 1][y2] - sum2[x2][y1 - 1] + sum2[x1 - 1][y1 - 1];
            }
            minn = min(minn, ans - res);
        }
        sort(a + 1, a + 1 + n, cmp3);
        for(int i = 1; i <= cnt; i++)
            for(int j = i + 1; j <= cnt; j++) {
            x1 = max(a[i].x1, a[j].x1);
            x2 = min(a[i].x2, a[j].x2);
            y1 = max(a[i].y1, a[j].y1);
            y2 = min(a[i].y2, a[j].y2);
            res = a[i].val1 + a[j].val1;
            if(x1 <= x2 && y1 <= y2) {
                res += sum2[x2][y2] - sum2[x1 - 1][y2] - sum2[x2][y1 - 1] + sum2[x1 - 1][y1 - 1];
            }
            minn = min(minn, ans - res);
        }
        printf("%d\n", minn);
    }
    return 0;
}