POJ2010--Moo University-Financial Aid（二分）

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题目大意：在C头牛里选N头牛，在不超过经费F的情况下，使得N头牛的得分中位数最大。

分析：首先，既然有两个属性，我们可以分别开两个数组，一个数组按成绩升序，一个数组按经费升序。二分中位数那头牛mid，然后按经费从少到多，添加mid两边的牛。再进行判断，四种情况，见代码。

代码：

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 222222;struct Cow{    int rank, score, aid;};Cow c_aid[maxn], c_score[maxn];int N, C, F;bool cmpa(Cow & i, Cow & j) {    return i.aid < j.aid;}bool cmps(Cow & i, Cow & j) {    return i.score < j.score;}int main() {    while(~scanf("%d%d%d", &N, &C, &F)) {        for(int i = 0; i < C; i++)            scanf("%d%d", &c_score[i].score, &c_score[i].aid);        sort(c_score, c_score+C, cmps);        for(int i = 0; i < C; i++)            c_score[i].rank = i;        memcpy(c_aid, c_score, sizeof(Cow)*C);        sort(c_aid, c_aid+C, cmpa);        int L = 0, R = C, ans = -1;        while(R-L > 1) {            int mid = (L+R)>>1;            int left = 0, right = 0, tot = c_score[mid].aid;            for(int i = 0; i < C; i++) {                if((c_aid[i].rank < mid) && (tot+c_aid[i].aid <= F) && (left < N/2)) {                    tot += c_aid[i].aid;                    left++;                }                else if((c_aid[i].rank > mid) && (tot+c_aid[i].aid <= F) && (right < N/2)) {                    tot += c_aid[i].aid;                    right++;                }            }            if(left < N/2 && right < N/2) {                ans = -1;                break;            }            else if(left < N/2) L = mid;            else if(right < N/2) R = mid;            else ans = c_score[mid].score, L = mid;        }        printf("%d\n", ans);    }    return 0;}

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