HDU 2489 Minimal Ratio Tree (dfs+Prim最小生成树)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2489

Problem Description

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

Given a complete graph of n nodes with all nodes and edges weighted, your task is to find a tree, which is a sub-graph of the original graph, with m nodes and whose 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 case contains two integers n (2<=n<=15) and m (2<=m<=n), which stands for the number of nodes in the graph and the number of nodes in the minimal ratio tree. Two zeros end the input. The next line
contains n numbers which stand for the weight of each node. The following n lines contain a diagonally symmetrical n×n connectivity matrix with each element shows the weight of the edge 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 of the matrix) are integers and in the range of [1, 100].

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

Output

For each test case output one line contains a sequence of the m nodes which constructs the minimal ratio tree. Nodes should be arranged in ascending order. If there are several such sequences, 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 are numbered 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

2008 Asia Regional Beijing

题意：

给出n个点，要从中选出m个点，要求选出的这m个点的所有边的边权值/点权值要最小！

并要输出所选的这m个点，如果有多种选择方法，那么就输出第一个点小的方案，如果第一个点相同就输出第二个点小的，一次类推！

PS：

由于这题的n比较小，只有15，所以可以先dfs枚举出所选择的点，然后在用最小生成树Prim算出最小的边权值的和；

代码如下：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std;
#define INF 1e18;
const double eps = 1e-9;
const int maxn = 17;
int n, m;
int e_val[maxn][maxn];
int node[maxn];
int ansn[maxn];//记录最终选得是哪些点
int tt[maxn];//记录中间过程选得是哪些点
int vis[maxn];
int low[maxn];
double minn;
int Prim(int s)
{
    int sum=0;
    memset(vis,0,sizeof(vis));
    for(int i = 1; i <= m; i++)
    {
        low[tt[i]] = e_val[s][tt[i]];
    }
    vis[s] = 1;
    low[s] = 0;
    int pos = s;
    for(int i = 1; i < m; i++)
    {
        int min_t = INF;
        for(int j = 1; j <= m; j++)
        {
            if(!vis[tt[j]] && min_t > low[tt[j]])
            {
                min_t = low[tt[j]];
                pos = tt[j];
            }
        }
        vis[pos] = 1;
        sum += min_t;
        for(int j = 1; j <= m; j++)
        {
            if(!vis[tt[j]] && e_val[pos][tt[j]] < low[tt[j]])
                low[tt[j]]=e_val[pos][tt[j]];
        }
    }
    return sum;
}
void DFS(int n_pre, int k)
{
    if(k == m)
    {
        double n_sum = 0;
        for(int i = 1; i <= m ; i++)
        {
            n_sum+=node[tt[i]];
        }
        double e_ans = 0;
        e_ans = Prim(tt[1]);
        double ans = e_ans/(n_sum*1.0);
        //if(ans < minn)
        if(ans - minn < -(eps))
        {
            minn = ans;
            for(int i = 1; i <= m; i++)
            {
                ansn[i] = tt[i];
            }
        }
        return ;
    }
    for(int i = n_pre+1; i <= n; i++)
    {
        tt[k+1] = i;
        DFS(i,k+1);
    }
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        if(n==0 && m==0)
            break;
        minn = INF;
        for(int i = 1; i <= n; i++)
        {
            scanf("%d",&node[i]);
        }
        for(int i = 1; i <= n; i++)
        {
            for(int j = 1; j <= n; j++)
            {
                scanf("%d",&e_val[i][j]);
            }
        }
        for(int i = 1; i <= n; i++)
        {
            tt[1] = i;
            DFS(i, 1);
        }
        for(int i = 1; i < m; i++)
        {
            printf("%d ",ansn[i]);
        }
        printf("%d\n",ansn[m]);
    }
    return 0;
}