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

3 230 20 100 6 26 0 32 3 02 21 10 22 00 0

 

Sample Output

1 31 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;}