HDU2489--Minimal Ratio Tree（最小生成樹）

Do more with less

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

#

根據上面的公式，找M個結點連成的權值最小的樹

思路

要找權值最小的m個連點連成的樹
給的數據說明圖中結點兩兩相連
所以只需要N個節點中枚舉M個，求出他們的最小生成樹，即可算出這M個結點最小權值
比較所以的權值即可得出最小

代碼

#include <iostream>
#include <cstdio>
#include <string>
#include <algorithm>
#include <cstring>
using namespace std;

const int INF = 1 << 25;
int n, m;
int map[30][30];
int wei[30];
int v[30];
int book[30];
int res[30];
double minall = INF;
void prim()
{
    int lowcost[30];
    int vis[30];
    int x = v[1];
    int sum = 0;
    memset(vis,0,sizeof(vis));
    for(int i = 1; i <= m; i ++)
        lowcost[v[i]] = map[x][v[i]];
    vis[x] = 1;
    for(int i = 1; i <= m; i ++)
    {
        int minx = INF;
        for(int j = 1; j <= m; j ++)
        {
            if(!vis[v[j]] && lowcost[v[j]] < minx)
            {
                x = v[j];
                minx = lowcost[v[j]];
            }
        }
        vis[x] = 1;
        if(minx != INF)
            sum += minx;  
        for(int j = 1; j <= m; j ++)
        {
            if(!vis[v[j]] && lowcost[v[j]] > map[x][v[j]])
                lowcost[v[j]] = map[x][v[j]];
        }
    }
    int sumn = 0;
    for(int i = 1; i <= m; i ++)
        sumn += wei[v[i]];
    double minthis = (sum*1.0) / (sumn*1.0);
    if(minthis < minall)
    {
        for(int i = 1; i <= m; i ++)
            res[i] = v[i];
        minall = minthis;
    }
}
void dfs(int x,int cnt)  
{  
    if(x == m + 1)
    {
        prim();
        return ;
    }
    for(int i = cnt;i <= n;i++)  
    {  
        if( !book[i] )  
        {  
            book[i] = 1;  
            v[x] = i;   
            dfs( x+1,i+1);  
            book[i] = 0;  
        }  
    }  
}  
int main()
{
    while(cin >> n >> m && n | m)
    {
        minall = INF;
        memset(book,0,sizeof(book));
        for(int i = 1; i <= n; i ++)
            cin >> wei[i];
        for(int i = 1; i <= n; i ++)
            for(int j = 1; j <= n; j ++)
                cin >> map[i][j];
        dfs(1,1);
        sort(res + 1, res + m + 1);
        for(int i = 1; i < m; i ++)
            cout << res[i] << " ";
        cout << res[m] << endl;
    }
    return 0;
}