HDU2489 Minimal Ratio Tree

Minimal Ratio Tree

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2570 Accepted Submission(s): 760

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个点的图和每个点之间的权值和点的权值，要求选出m个点，使得m个点之间边的权值之和比上点的权值之和最小。。。

首先暴力DFS，选出m个点，然后对这m个点做最小生成树就可以了。。。

这题思路很简单，但是还是WA了很多发。。。这题和POJ3925一样，但是POJ上比较两个浮点数直接比较就可以过，HDOJ上比较要考虑精度问题，不考虑会WA。。

#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>using namespace std;const int MAXN=20;const int INF=1<<30;const double eps=1e-5;int lowcost[MAXN],w[MAXN][MAXN],ch[MAXN],node[MAXN];int mp[MAXN][MAXN];bool vis[MAXN];int n,m;int prim(int v0){    int i,j,mincost,minone;    memset(vis,0,sizeof(vis));    int sum=0;    for(i=1;i<=m;i++)    {        lowcost[i]=mp[v0][i];    }    lowcost[v0]=0;    vis[v0]=1;    for(i=1;i<=m-1;i++)    {        mincost=INF;        for(j=1;j<=m;j++)        {            if(!vis[j]&&mincost>lowcost[j])            {                mincost=lowcost[j];                minone=j;            }        }        vis[minone]=1;        sum+=mincost;        for(j=1;j<=m;j++)        {            if(!vis[j]&&mp[minone][j]<lowcost[j])            {                lowcost[j]=mp[minone][j];            }        }    }    return sum;}double minans;int ans[MAXN];void dfs(int k,int num){    int i,j;    if(num==m)    {        double tot=0;        for(i=1;i<=m;i++)            tot+=(double)node[ch[i]];        for(i=1;i<=m;i++)        {            for(j=1;j<=m;j++)                mp[i][j]=w[ch[i]][ch[j]];        }        double sum=prim(1)*1.0;        double temp=sum/tot;        if(minans-temp>=eps)        {            minans=temp;            for(i=1;i<=m;i++)            {                ans[i]=ch[i];            }        }        return;    }    for(i=k+1;i<=n;i++)    {        ch[num+1]=i;        dfs(i,num+1);    }}int main(){    int i,j;    while(scanf("%d%d",&n,&m)==2)    {        if(n==0&&m==0)            break;        for(i=1;i<=n;i++)            scanf("%d",&node[i]);        for(i=1;i<=n;i++)        {            for(j=1;j<=n;j++)            {                scanf("%d",&w[i][j]);            }        }        minans=INF;        for(i=1;i<=n;i++)        {            ch[1]=i;            dfs(i,1);        }        printf("%d",ans[1]);        for(i=2;i<=m;i++)            printf(" %d",ans[i]);        printf("\n");    }    return 0;}