poj 2010

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优先队列

先将奶牛排序，考虑每个奶牛作为中位数时，比它分数低（前面的）的那群牛的学费总和left，后面的总和right。然后从分数高往分数低扫描，满足left + y + right <= F的第一个解就是最优解。

#include <iostream>#include <queue>#include <algorithm>#include <cstring>using namespace std;const int MAXN = 100005;struct P {int x;int y;int left;int right;}e[MAXN];bool cmp(P a, P b) {return a.x < b.x;}int main() {int n, c, f;memset(e, 0, sizeof(e));scanf("%d%d%d", &n, &c, &f);for(int i=1; i<=c; i++)scanf("%d%d", &e[i].x, &e[i].y);sort(e+1, e+c+1, cmp);if(n > 1) {priority_queue<int> que;int p, t, sum=0;for(int i=1; i<=n/2; i++) {p = e[i].y;que.push(p);sum += p;}for(int i=n/2+1; i<=c-n/2; i++) {e[i].left = sum;p = que.top();t = e[i].y;if(t < p) {que.pop();que.push(t);sum = sum - p + t;}}while(que.size())que.pop();sum = 0;for(int i=c; i>c-n/2; i--) {p = e[i].y;que.push(p);sum += p;}for(int i=c-n/2; i>n/2; i--) {e[i].right = sum;p = que.top();t = e[i].y;if(t < p) {que.pop();que.push(t);sum = sum - p + t;}}int ans = -1;for(int i=c-n/2; i>n/2; i--) {int total = e[i].left + e[i].y + e[i].right;if(total <= f) {ans = e[i].x;break;}}printf("%d\n", ans);}return 0;}

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