POJ2010 Moo University

　　题目地址

　　分析：如果用二分法，关键是score和aid分开排序，score排序是为了充分利用中位数的性质，这样就可以确定m左右必须各选N/2个，到这之后有人是用dp求最优解，可以再次按照aid排序一次，可以直接确定最优解（肯定是从最小的开始选择！）：

#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
int N, C, F;
const int maxn = 100010;
struct Cow{
    int id, score, aid;
}Score[maxn], Aid[maxn];
bool cmp_score( const Cow &a, const Cow &b) {
    return a.score < b.score;
}
bool cmp_aid(const Cow &a, const Cow &b) {
    return a.aid < b.aid;
}
int main(void) {

    scanf("%d%d%d", &N, &C, &F);
    for (int i = 0; i < C; i++) scanf("%d%d", &Score[i].score, &Score[i].aid);
    sort(Score, Score+C, cmp_score);
    for (int i = 0; i < C; i++) Score[i].id = i;
    memcpy(Aid, Score, sizeof(Cow)*C);
    sort(Aid, Aid+C, cmp_aid);
    int l = 0, u = C, ans = -1;
    while (u-l > 1) {
        int m = (u+l) >> 1;
        int left = 0, right = 0, total = Score[m].aid;
        for (int i = 0; i < C; i++) {
            if (Aid[i].id < m && (total+Aid[i].aid) <= F && left < N/2) {
                total += Aid[i].aid;
                left++;
            } else if (Aid[i].id > m && (total+Aid[i].aid) <= F && right < N/2) {
                total += Aid[i].aid;
                right++;
            }
        }
        if (left < N/2 && right < N/2) {//insufficient
            ans = -1;
            break;
        } else if (left < N / 2) {
                l = m;
        } else if (right < N / 2){
                u = m;
        } else {
            ans = Score[m].score;
            l = m;
        }
    }        
    printf("%d
", ans);
    return 0;
}

　　百度还看到用堆、优先队列做的！mark。　　

　　参考：

　　　　《挑战程序设计竞赛》3.1