poj_3925 Minimal Ratio Tree

Minimal Ratio Tree

Time Limit: 1000MS

Memory Limit: 65536K

Total Submissions: 836

Accepted: 304

题目链接：http://poj.org/problem?id=3925

Description

For a tree, which nodes and edges are all weighted, the ratio of it iscalculated according to the following equation.

Given a complete graph of n nodes with all nodes and edges weighted, your taskis to find a tree, which is a sub-graph of the original graph, with m nodes andwhose ratio is the smallest among all the trees of m nodes in the graph.

Input

Input contains multiple test cases. The first line of each test casecontains two integers n (2<=n<=15) and m (2<=m<=n), which standsfor the number of nodes in the graph and the number of nodes in the minimalratio tree. Two zeros end the input. The next line contains n numbers whichstand for the weight of each node. The following n lines contain a diagonallysymmetrical n×n connectivity matrix with each element shows the weight of theedge connecting one node with another. Of course, the diagonal will be all 0,since there is no edge connecting a node with itself.

All the weights of both nodes and edges (except for the ones on the diagonal ofthe matrix) are integers and in the range of [1, 100].

The figure below illustrates the first test case in sample input. Node 1 andNode 3 form the minimal ratio tree.

Output

For each test case output oneline contains a sequence of the m nodes which constructs the minimal ratiotree. Nodes should be arranged in ascending order. If there are several suchsequences, pick the one which has the smallest node number; if there's a tie,look at the second smallest node number, etc. Please note that the nodes arenumbered from 1.

Sample Input

3 2

30 20 10

0 6 2

6 0 3

2 3 0

2 2

1 1

0 2

2 0

0 0

Sample Output

1 3

1 2

Source

Beijing 2008

 

题意：

         给你一个完全图，求其中m个点中的radio最小radio=（每条边的权值之和）/（每个点的权值之和）

解题思路：

         我是先求出n个点中m个点的每一种组合，在每一种组合中的点，在用prim算法求这些点的最小生成树的值，再计算该组合的radio值，并保存最小的radio对应的点，最后输出每个点

代码：

#include <iostream>#include<cstdio>#include<cstring>#define maxn 100#define VALUE 0xffffffusing namespace std;struct edgeRadio{int member[maxn];int t;};int noteweight[maxn];//点权int edgeweight[maxn][maxn];//边权edgeRadio edgeradio;int n,m;int visited[maxn];int minCost[maxn];double minValue;edgeRadio minEdge;//prim算法求最小值int prim(){int i;int res=0;for(i=0;i<=n;i++){visited[i]=0;minCost[i]=VALUE;}minCost[edgeradio.member[0]]=0;while(true){int t=-1;for(i=1;i<=m;i++){if(visited[edgeradio.member[i-1]]==0 && (t==-1 || minCost[edgeradio.member[i-1]]<minCost[t]))t=edgeradio.member[i-1];}if(t==-1)break;visited[t]=1;res+=minCost[t];for(i=1;i<=m;i++){if(minCost[edgeradio.member[i-1]]>edgeweight[edgeradio.member[i-1]][t] && edgeweight[edgeradio.member[i-1]][t]!=0){minCost[edgeradio.member[i-1]]=edgeweight[edgeradio.member[i-1]][t];}}}return res;}//求radiodouble getRadio(){int num=0;int res=prim();for(int i=1;i<=m;i++){num+=noteweight[edgeradio.member[i-1]];}return res*1.0/num;}//递归求每一种组合void getNum(int t,int p){if(t==0){double res = getRadio();if(minValue>res){minValue=res;minEdge=edgeradio;}return ;}else{for(int i=p;i<=n;i++){edgeradio.member[edgeradio.t]=i;edgeradio.t++;getNum(t-1,p+1);edgeradio.t--;p++;}}}int main(){    int i,j;    while(true)    {        scanf("%d%d",&n,&m);        minValue=VALUE;        if(n==0 && m==0)            break;memset(edgeweight,0,sizeof(edgeweight));        for(i=1;i<=n;i++)        {            scanf("%d",¬eweight[i]);        }        for(i=1;i<=n;i++)        {            for(j=1;j<=n;j++)            {                scanf("%d",&edgeweight[i][j]);            }        }getNum(m,1);for(i=0;i<minEdge.t-1;i++){printf("%d ",minEdge.member[i]);}printf("%d\n",minEdge.member[i]);    }    return 0;}