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 0xffffff
using 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radio
double 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;
}