POJ - 3111 K Best 平均值最大（01分数规划）

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题目大意：有n颗珍珠，每颗珍珠有相应的价值和代价，要从中挑出K颗珍珠，使得（这k颗珍珠的总价值/这k颗珍珠的代价）达到最大

解题思路：平均值最大，要注意精度
给出01分数规划的详细解读
< http://www.cnblogs.com/perseawe/archive/2012/05/03/01fsgh.html>

#include<cstdio>#include<algorithm>#include<cstring>using namespace std;#define maxn 100010int n, k, ans[maxn];double Sum;struct jewels {    double v, w, d;    int id;}J[maxn];bool cmp(const jewels a, const jewels b) {    return a.d > b.d;}bool judge(double mid) {    for(int i = 0; i < n; i++)        J[i].d = J[i].v - mid * J[i].w;    sort(J, J + n, cmp);    double t = 0.0;    for(int i = 0; i < k; i++)        t += J[i].d;    return t >= 0;}void solve() {    double l = 0, r = 1e7 + 1;    while(r - l > 1e-6) {        double mid = (l + r) / 2;        if(judge(mid))            l = mid;        else            r = mid;    }}int main() {    while(scanf("%d%d", &n, &k) != EOF) {        for(int i = 0; i < n; i++) {            scanf("%lf%lf", &J[i].v, &J[i].w);            J[i].id = i + 1;        }        solve();        printf("%d", J[0].id);        for(int i = 1; i < k; i++)            printf(" %d", J[i].id);        printf("\n");    }    return 0;}

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