HDU 6296 代碼派對 (二維前綴和 差分)

題目鏈接:http://acm.hdu.edu.cn/showproblem.php?pid=6296

利用二維前綴和，我們可以求出每個位置sumv[i][j]被矩形覆蓋的次數，然後在覆蓋矩陣數裏面排列組合選出3個C[sum[i][j]]

但裏面會有重複。利用差分的思想，減掉去除左邊,去除上邊的前綴和，以及再加一次左上的前綴和(因爲被減了兩次)

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <map>
#include <set>
#include <vector>
#include <string>
#include <cstring>
using namespace std;
typedef long long ll;
static const int MAX_N = 1e5 + 5;
static const ll Mod = 233;
static const int N = 105;
static const int INF = 0x3f3f3f3f;
ll C[MAX_N], sumv[1005][1005];
int re[MAX_N][4];
ll solve(int dx, int dy, int n){
    memset(sumv, 0, sizeof(sumv));
    for(int i = 0; i < n; ++i){
        int x1 = re[i][0], y1 = re[i][1], x2 = re[i][2] + dx, y2 = re[i][3] + dy;
        sumv[x1][y1]++;
        sumv[x1][y2 + 1]--;
        sumv[x2 + 1][y1]--;
        sumv[x2 + 1][y2 + 1]++;    //二維前綴和更新1
    }
    ll ret = 0;
    for(int i = 1; i <= 1000; ++i){
        for(int j = 1; j <= 1000; ++j){
            sumv[i][j] += sumv[i - 1][j] + sumv[i][j - 1] - sumv[i - 1][j - 1];   
                 //二維前綴和
            ret += C[sumv[i][j]];
        }
    }
    return ret;
}
int main(){
//    freopen("input.txt", "r", stdin);
//    freopen("output.txt", "w", stdout);
    int T;
    C[0] = C[1] = C[2] = 0;
    for(int i = 3; i <= 100000; ++i){
        C[i] = (ll)i * (i - 1) * (i - 2) / 6;
        //printf("%lld\n", C[i]);
    }
    scanf("%d", &T);
    while(T--){
        int n;
        scanf("%d", &n);
        for(int i = 0; i < n; ++i){
            for(int j = 0; j < 4; ++j) scanf("%d", &re[i][j]);
        }
        printf("%lld\n", solve(0, 0, n) - solve(0, -1, n) - solve(-1, 0, n) + solve(-1, -1, n));
    }
    return 0;
}