AcWing - 513 - 无线网络发射器选址 = 前缀和

Posted 2021-03-26 inko

https://www.acwing.com/problem/content/515/

二维前缀和暴力统计，注意最后放这些点的位置的时候可以放在角落里的，巨坑，应该最简单的思路是枚举每个点1~MAXN，然后写个函数自动返回他周围的点的和。

#include <bits/stdc++.h>
using namespace std;

const int MAXN = 129;

int sum[MAXN + 25][MAXN + 25];

int main() {
#ifdef Yinku
    freopen("Yinku.in", "r", stdin);
#endif // Yinku
    int d, n;
    scanf("%d%d", &d, &n);
    d *= 2;
    while(n--) {
        int x, y, k;
        scanf("%d%d%d", &x, &y, &k);
        ++x, ++y;
        sum[x][y] += k;
    }

    for(int i = 1; i <= MAXN + d / 2; ++i) {
        for(int j = 1; j <= MAXN + d / 2; ++j) {
            sum[i][j] += sum[i - 1][j] + sum[i][j - 1] - sum[i - 1][j - 1];
            //printf("%2d", sum[i][j]);
        }
        //printf("
");
    }

    int ans2 = 0;
    for(int i = d + 1;  i <= MAXN + d / 2; ++i) {
        for(int j = d + 1 ; j <= MAXN + d / 2; ++j) {
            ans2 = max(ans2, sum[i][j] - sum[i - (d + 1)][j] - sum[i][j - (d + 1)] + sum[i - (d + 1)][j - (d + 1)]);
            //printf("%2d", sum[i][j] - sum[i - (d + 1)][j] - sum[i][j - (d + 1)] + sum[i - (d + 1)][j - (d + 1)]);
        }
        //printf("
");
    }
    int ans1 = 0;
    for(int i = d / 2 + 1;  i <= MAXN + d / 2; ++i) {
        for(int j = d / 2 + 1 ; j <= MAXN + d / 2; ++j) {
            if(sum[i][j] - ((i >= d + 1) ? sum[i - (d + 1)][j] : 0) - ((j >= d + 1) ? sum[i][j - (d + 1)] : 0) + ((i >= d + 1 && j >= d + 1) ? sum[i - (d + 1)][j - (d + 1)] : 0) == ans2)
                ++ans1;
        }
    }
    printf("%d %d
", ans1, ans2);
    return 0;
}