POJ 2728 最优比率生成树

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题目LINK: http://poj.org/problem?id=2728
原理参考： http://www.cnblogs.com/lotus3x/archive/2009/03/21/1418480.html
http://hi.baidu.com/zzningxp/item/28aa46e0fd86bdc2bbf37d03

做法就是，二分这个比率r, 把生成树的边的权值设为 cost[i] - r * len[i] , 求最小生成树结果。如果结果为0，则是最有比率，否则不断二分（单调递减的函数）

//minimum ratio of overall cost of the channels to the total length.#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>using namespace std;#define INF 1e10#define eps 0.00001#define N 1111struct point{    double x, y, h;}p[N];double cost[N][N], len[N][N], dis[N][N], dd[N];int n;bool vis[N];double get_dis(int i, int j){    return sqrt((p[i].x - p[j].x) * (p[i].x - p[j].x) + (p[i].y - p[j].y) * ( p[i].y - p[j].y));}double get_h(int i, int j){    return fabs(p[i].h - p[j].h);}double sol(double in){    for(int i=1; i<=n; i++)        for(int j=1; j<=n; j++)            dis[i][j] = cost[i][j] - in * len[i][j];     double ret = 0;    memset(vis, 0, sizeof(vis));    for(int i=1; i<=n; i++) dd[i] = INF;    dd[1] = 0;    for(int i=1; i<=n; i++)    {        double mi = INF;        int id = -1;        for(int j=1; j<=n; j++)        {            if(vis[j]) continue;            if(dd[j] < mi)            {                mi = dd[j] ; id = j;            }        }        vis[id ] = 1;        ret += mi;        for(int j=1; j<=n; j++)        {            if(vis[j] ) continue;            if(dd[j] > dis[id][j]) dd[j] = dis[id][j] ;        }    }    return ret;}int main(){#ifndef ONLINE_JUDGE    freopen("in.txt", "r", stdin);#endif // ONLINE_JUDGE    while(scanf("%d", &n) && n)    {        for(int i=1; i<=n; i++)        {            scanf("%lf%lf%lf", &p[i].x, &p[i].y, &p[i].h);        }        for(int i=1; i<=n; i++)        {            for(int j=i; j<=n; j++)            {                if(i == j) cost[i][j] = len [i][j] = 0;                else                {                    len[i][j] = len[j][i] = get_dis(i, j);                    cost[i][j] = cost[j][i] = get_h(i, j);                }            }        }        double l = 0, r = 10001111;// r == 1000 竟然也可以跑过！        while(fabs(l - r) > eps)        {            double m = (l+r)/2;            double tmp = sol(m);            if(tmp > eps) l= m;            else r = m;        }        printf("%.3lf\n", l);    }    return 0;}

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