POJ 2728 Desert King（最优比率生成树）

题意：

每条路径有一个 cost 和 dist，求图中 sigma(cost) / sigma(dist) 最小的生成树。

思路：

1. sigma(cost)/sigma(dist) <= r 有 cost - r * dist <= 0

2. r = 0 当然是理想的初始状态，我们不断尝试 r 使结果不断逼近最优值，方能得到结果。

 

#include <iostream>

#include <cmath>

#include <algorithm>

using namespace std;



const int MAXN = 1010;

const double INFS = 1e15;

const double eps = 1e-5;



double x[MAXN], y[MAXN], z[MAXN];

double dist[MAXN][MAXN], cost[MAXN][MAXN], map[MAXN][MAXN], d[MAXN];

int p[MAXN], n;

bool vis[MAXN];



double prim(double ratio) {

    for (int i = 1; i <= n; i++) {

        map[i][i] = 0.0;

        for (int j = i+1; j <= n; j++)

            map[i][j] = map[j][i] = cost[i][j] - ratio*dist[i][j];

    }

    for (int i = 1; i <= n; i++)

        d[i] = map[1][i], vis[i] = false, p[i] = 1;

    vis[1] = true;

    for (int k = 1, j; k < n; k++) {

        double mind = INFS;

        for (int i = 2; i <= n; i++) 

            if (!vis[i] && mind > d[i])

                mind = d[i], j = i;

        vis[j] = true;

        for (int i = 2; i <= n; i++) 

            if (!vis[i] && d[i] > map[j][i])

                d[i] = map[j][i], p[i] = j;

    }

    double d1 = 0, c1 = 0;

    for (int i = 2; i <= n; i++) {

        d1 += dist[i][p[i]];

        c1 += cost[i][p[i]];

    }

    return c1 / d1;

}



int main() {

    while (scanf("%d", &n) && n) {

        for (int i = 1; i <= n; i++)

            scanf("%lf%lf%lf", &x[i], &y[i], &z[i]);

        for (int i = 1; i <= n; i++) {

            dist[i][i] = cost[i][i] = 0.0;

            for (int j = i+1; j <= n; j++) {

                dist[i][j] = dist[j][i] = sqrt((x[i]-x[j])*(x[i]-x[j]) + (y[i]-y[j])*(y[i]-y[j]));

                cost[i][j] = cost[j][i] = fabs(z[i] - z[j]);

            }

        }

        double r1 = 0.0, r2 = 0.0;

        while (true) {

            r2 = prim(r1);

            if (fabs(r1 - r2) < eps)

                break;

            r1 = r2;

        }

        printf("%.3lf\n", r1);

    }

    return 0;

}