BNUOJ 13268 Aladdin and the Optimal Invitation

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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;}

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