hdu 2489 Minimal Ratio Tree【Dfs+kruskal】

Minimal Ratio Tree

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3928    Accepted Submission(s): 1207

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

 

 思路：

因为n比较小，所以我们可以采取直接暴力Dfs枚举所有可能选取的点的方案，然后累加点权值，然后将这些点相关的边加入数组中，贪心求最小生成树即可。

注意的点：我们在Dfs的时候，如果每一层都遍历n个点，是会超时的，所以我们枚举当前点是否加入子图中，这样就变成了每层枚举两个操作，防止超时的情况。

AC代码：

#include<string.h>
#include<algorithm>
#include<stdio.h>
using namespace std;
struct zuobiao
{
    int x,y,w;
}a[121212];
int d[250];
int map[250][250];
int f[250];
int vis[250];
int output[250];
int n,m;
double minn;
int cmp(zuobiao a,zuobiao b)
{
    return a.w<b.w;
}
int find(int a)
{
    int r=a;
    while(f[r]!=r)
    r=f[r];
    int i=a;
    int j;
    while(i!=r)
    {
        j=f[i];
        f[i]=r;
        i=j;
    }
    return r;
}
void merge(int a,int b)
{
    int A,B;
    A=find(a);
    B=find(b);
    if(A!=B)
    f[B]=A;
}
void solve(int sumd)
{
    int tot=0,sume=0;
    for(int i=0;i<n;i++)
    {
        f[i]=i;
    }
    for(int i=0;i<n;i++)
    {
        if(vis[i]==1)
        {
            for(int j=0;j<n;j++)
            {
                if(vis[j]==0)continue;
                if(map[i][j]==0)continue;
                a[tot].x=i;
                a[tot].y=j;
                a[tot].w=map[i][j];
                tot++;
            }
        }
    }
    sort(a,a+tot,cmp);
    for(int i=0;i<tot;i++)
    {
        if(find(a[i].x)!=find(a[i].y))
        {
            merge(a[i].x,a[i].y);
            sume+=a[i].w;
        }
    }
    int tot2=0;
    double out=sume*1.0/sumd;
    if(out<minn||minn==-1)
    {
        minn=out;
        for(int i=0;i<n;i++)
        {
            if(vis[i]==1)
            output[tot2++]=i;
        }
    }
}
void Dfs(int cont,int now,int sumd)
{
    if(now>n)return ;
    if(cont==m)
    {
        solve(sumd);
        return ;
    }
    vis[now]=1;
    Dfs(cont+1,now+1,sumd+d[now]);
    vis[now]=0;
    Dfs(cont,now+1,sumd);
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        if(n==0&&m==0)break;
        for(int i=0;i<n;i++)
        {
            scanf("%d",&d[i]);
        }
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<n;j++)
            {
                scanf("%d",&map[i][j]);
            }
        }
        int ans[25];
        memset(vis,0,sizeof(vis));
        minn=-1;
        Dfs(0,0,0);
        for(int i=0;i<m;i++)
        {
            if(i==0)
            printf("%d",output[i]+1);
            else printf(" %d",output[i]+1);
        }
        printf("\n");
    }
}
/*
3 2
30 10 20
0 3 1
3 0 1
1 1 0
//////1 3
3 2
30 30 10
0 3 1
3 0 1
1 1 0
//////1 3
3 2
30 80 10
0 3 1
3 0 1
1 1 0
//////2 3
*/