BNUOJ 13268 Aladdin and the Optimal Invitation

思路：x轴和y轴的情况是独立的，可以分开考虑，那么只要枚举位置，能维护最小值即可，这只要把公式拆开，预处理出两个前缀和即可，一个是w的前缀和，一个是w * x的前缀和

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int N = 50005;

typedef long long ll;
int t, n, m, q;

int x[N], y[N];
ll sum[N][2], w[N];

ll cal(int v, int n) {
    ll ans1 = sum[v][1] * v - sum[v][0];
    ll ans2 = (sum[n][0] - sum[v][0]) - (sum[n][1] - sum[v][1]) * v;
    return ans1 + ans2;
}

int gao(int *x, int n) {
    memset(sum, 0, sizeof(sum));
    for (int i = 0; i < q; i++) {
        sum[x[i]][0] += w[i] * x[i];
        sum[x[i]][1] += w[i];
    }
    for (int i = 1; i <= n; i++) {
        for (int j = 0; j < 2; j++)
            sum[i][j] += sum[i - 1][j];
    }
    int ans = 1;
    for (int i = 1; i <= n; i++) {
        if (cal(i, n) < cal(ans, n))
            ans = i;
    }
    return ans;
}

int main() {
    int cas = 0;
    scanf("%d", &t);
    while (t--) {
        scanf("%d%d%d", &n, &m, &q);
        for (int i = 0; i < q; i++)
            scanf("%d%d%lld", &x[i], &y[i], &w[i]);
        printf("Case %d: %d %d\n", ++cas, gao(x, n), gao(y, m));
    }
    return 0;
}

BNUOJ 13268 Aladdin and the Optimal Invitation

原文：http://blog.csdn.net/accelerator_/article/details/44735085