hdu Minimal Ratio Tree（最小生成树---prim）

Minimal Ratio Tree

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1285    Accepted Submission(s): 383

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

 

Recommend

gaojie

因为数据范围很小，可以枚举所有点集的情况，求最小的比例

#include <stdio.h>
#include <string.h>


int n,m;
int Nodep[20];
bool chose[20];
int map[20][20];
int ans[20];
double minn;
bool vist[20];
int dis[20];

double Prim(int f){
    int i,j,x;
    memset(vist,0,sizeof(vist));
    for(i = 0;i < 20;i++)
        dis[i] = 10000000;
    dis[f] = 0;
    int minx,min,ansx = 0;;
    for(i = 1;i <= m;i++){//把m写成n导致改了很久
        min = 100000000;
        for(j = 1;j <= n;j++)
            if(chose[j] && !vist[j] && dis[j] < min){
                min = dis[j];
                minx = j;
            }
        vist[minx] = 1;
        ansx += dis[minx];
        dis[minx] = 0;//加上这一句会更保险，比如这里即使m写成n了，也不会出错
        for(j = 1;j <= n;j++)
            if(chose[j] && !vist[j] && map[minx][j] < dis[j])
                dis[j] = map[minx][j];
    }
    return ansx * 1.0;
}

void Dfs(int p,int c){
    if(c == m){
        int first = 0;;
        int countn = 0;
        double tem,tem1,tem2 = 0;

        for(int i = 1;i <= n;i++)
            if(chose[i]){
                if(!first){
                    first = i;
                    tem1 = Prim(first);
                }
                tem2 += Nodep[i] * 1.0;
            }

        tem = tem1 / tem2;
        if(tem <  minn){
            minn = tem;
            for(int i = 1;i <= n;i++)
                if(chose[i]){
                    ans[countn] = i;
                    countn++;
                }
        }
        return;
    }
    chose[p] = 1;
    Dfs(p+1,c+1);
    chose[p] = 0;
    if(n - p + c >= m)
        Dfs(p+1,c);
}

int main(){
    int i,j;

    while(scanf("%d%d",&n,&m) && n && m){
        memset(chose,0,sizeof(chose));
        minn = 1000;
        for(i = 1;i <= n;i++)
            scanf("%d",&Nodep[i]);
        for(i = 1;i <= n;i++)
            for(j = 1;j <= n;j++)
                scanf("%d",&map[i][j]);
        Dfs(1,0);
        for(i = 0;i < m - 1;i++)
            printf("%d ",ans[i]);
        printf("%d\n",ans[m-1]);
    }
    return 0;
}