POJ 2728 Desert King 最优比率生成树

Desert King

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 20978		Accepted: 5898

【Description】

David the Great has just become the king of a desert country. To win the respect of his people, he decided to build channels all over his country to bring water to every village. Villages which are connected to his capital village will be watered. As the dominate ruler and the symbol of wisdom in the country, he needs to build the channels in a most elegant way.
After days of study, he finally figured his plan out. He wanted the average cost of each mile of the channels to be minimized. In other words, the ratio of the overall cost of the channels to the total length must be minimized. He just needs to build the necessary channels to bring water to all the villages, which means there will be only one way to connect each village to the capital.
His engineers surveyed the country and recorded the position and altitude of each village. All the channels must go straight between two villages and be built horizontally. Since every two villages are at different altitudes, they concluded that each channel between two villages needed a vertical water lifter, which can lift water up or let water flow down. The length of the channel is the horizontal distance between the two villages. The cost of the channel is the height of the lifter. You should notice that each village is at a different altitude, and different channels can't share a lifter. Channels can intersect safely and no three villages are on the same line.
As King David's prime scientist and programmer, you are asked to find out the best solution to build the channels.

【Input】

There are several test cases. Each test case starts with a line containing a number N (2 <= N <= 1000), which is the number of villages. Each of the following N lines contains three integers, x, y and z (0 <= x, y < 10000, 0 <= z < 10000000). (x, y) is the position of the village and z is the altitude. The first village is the capital. A test case with N = 0 ends the input, and should not be processed.

【Output】

For each test case, output one line containing a decimal number, which is the minimum ratio of overall cost of the channels to the total length. This number should be rounded three digits after the decimal point.

【Sample Input】

4
0 0 0
0 1 1
1 1 2
1 0 3
0

【Sample Output】

1.000

【题意】

给出一张完全图，每条边都有长度和花费，要求在图中找到一棵生成树，使得Sum(Cost)/Sum(dist)达到最小。

【分析】

据说05年ACM的某场比赛上，教主怒切一题最优比率生成树，坑死了无数跟榜着...-_-////

 

最优比率生成树的前导知识是01分数规划。

基本思路是Dinkelbach逼近法：

整体思路跟原本的01分数规划基本相同，方程F(L)=Sum(cost[i])/Sum(dist[i])，只要把L'的生成过程改成Prim即可。

Prim堆加边的时候，用cost-l*dist作为边权。

/* ***********************************************
MYID    : Chen Fan
LANG    : G++
PROG    : PKU_2728
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <bitset>

using namespace std;

typedef struct nod
{
    int x,y,z;
} node;
node p[1010];

double getdist(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));
}

typedef struct enod
{
    int x,y;
    double dist,cost,r;
    friend bool operator < (enod a,enod b)
    {
        return a.r>b.r;
    }
} enode;

enode gete(int x,int y,double dist,double cost,double l)
{
    enode a;
    a.x=x;a.y=y;a.dist=dist;a.cost=cost;
    a.r=cost-l*dist;
    return a;
}

double prim(int n,double l)
{
    priority_queue<enode> q;
    while (!q.empty()) q.pop();
    bitset<1010> flag;
    flag.reset();
    flag[1]=1;
    double cost=0,dist=0;
    for (int i=2;i<=n;i++) q.push(gete(1,i,getdist(1,i),abs(p[1].z-p[i].z),l));

    for (int i=1;i<n;i++)
    {
        enode now=q.top();
        q.pop();
        while (flag[now.y])
        {
            now=q.top();
            q.pop();
        }
        flag[now.y]=1;
        cost+=now.cost;dist+=now.dist;
        for (int j=1;j<=n;j++)
        if (j!=now.y&&!flag[j])
        q.push(gete(now.y,j,getdist(now.y,j),abs(p[now.y].z-p[j].z),l));
    }    

    return cost/dist;
}

int main()
{
    freopen("2728.txt","r",stdin);
    
    int n;
    while (scanf("%d",&n))
    {
        if (n==0) break;

        for (int i=1;i<=n;i++) scanf("%d%d%d",&p[i].x,&p[i].y,&p[i].z);

        double l=1,ans;
        while (1)
        {
            ans=prim(n,l);
            if (fabs(ans-l)<1e-6) break;
            else l=ans;
        }

        printf("%.3f\n",ans);
    }

    return 0;
}