LuoguP1369 矩形 前缀和

分别预处理每行每列前缀合，然后枚举两顶点坐标即可。读取线段上值时要注意去重！

#include<iostream>
#include<cstdio>
using namespace std;

const int Maxn = 110;
int xs[Maxn][Maxn],ys[Maxn][Maxn],g[Maxn][Maxn];
int n,x,y,ans;

int tot(int x1,int y1,int x2,int y2){
    if(x1 == x2)return ys[x2][y2]-ys[x2][y1-1];
    if(y1 == y2)return xs[x2][y2]-xs[x1-1][y1];
    return xs[x2][y1]-xs[x1-1][y1]+xs[x2][y2]-xs[x1-1][y2]
        +ys[x1][y2-1]-ys[x1][y1]+ys[x2][y2-1]-ys[x2][y1];
}

void print(int a[110][110]){
    for(int i = 1;i <= 11;i++){
        for(int j = 1;j <= 11;j++)printf("%d ",a[i][j]);
        putchar('
');
    }
    putchar('
');
}

int main(){
    cin >> n;
    for(int i = 1;i <= n;i++){
        cin >> x >> y;
        g[x][y] = 1;
    }
    for(int i = 1;i <= 100;i++){
        ys[i][0] = xs[0][i] = 0;
        for(int j = 1;j <= 100;j++){
            ys[i][j] = ys[i][j-1]+g[i][j];
            xs[j][i] = xs[j-1][i]+g[j][i];
        }
    }
    for(int x1 = 1;x1 <= 100;x1++)
        for(int y1 = 1;y1 <= 100;y1++)
            for(int x2 = x1;x2 <= 100;x2++)
                for(int y2 = y1;y2 <= 100;y2++)
                    ans = max(ans,tot(x1,y1,x2,y2));
    cout << ans;
return 0;
}

 ps：好像还可以用一般的二维前缀和外层减里层处理。。