POJ2728 Desert King 题解 最优比率生成树 01分数规划

题目链接：http://poj.org/problem?id=2728

题目大意：

给你一个无向图，每条边有一个价值和长度，求一棵生成树，其对应的价值和与长度和的币值最小。

解题思路：
01分数规划。对每一个 (mid)，以 (a_i - mid imes b_i) 为边求最小生成树，判断最小生成树的长度是否 (ge 0)。

示例代码：

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <iostream>
using namespace std;
const int maxn = 1010;
int n, x[maxn], y[maxn], z[maxn];
double dis[maxn][maxn], cost[maxn];
bool vis[maxn];
bool prim(double mid) {
    memset(vis, 0, sizeof(bool)*n);
    for (int i = 0; i < n; i ++) {
        for (int j = 0; j < i; j ++) {
            dis[i][j] = dis[j][i] = abs(z[i] - z[j]) - mid * sqrt((x[i]-x[j])*(x[i]-x[j]) + (y[i]-y[j])*(y[i]-y[j]));
        }
    }
    cost[0] = 0;
    for (int i = 1; i < n; i ++) cost[i] = 1e9;
    double ans = 0;
    for (int j = 0; j < n; j ++) {
        int u = -1;
        for (int i = 0; i < n; i ++) if (!vis[i] && (u == -1 || cost[u] > cost[i])) u = i;
        if (u == -1) break;
        ans += cost[u];
        vis[u] = true;
        for (int i = 0; i < n; i ++) if (!vis[i] && dis[u][i] < cost[i]) cost[i] = dis[u][i];
    }
    return ans >= 0;
}
int main() {
    while (~scanf("%d", &n) && n) {
        for (int i = 0; i < n; i ++) scanf("%d%d%d", x+i, y+i, z+i);
        double L = 0, R = 100.0;
        while (R - L > 1e-4) {
            double mid = (L + R) / 2;
            if (prim(mid)) L = mid;
            else R = mid;
        }
        printf("%.3f
", L);
    }
    return 0;
}