Minimal Ratio Tree

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 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. 

3 2
30 20 10
0 6 2
6 0 3
2 3 0
2 2
1 1
0 2
2 0
0 0

1 3
1 2

给你一张图n个点，每个点有权值，问你选出m个点，使得最小，输出方案。

解题思路：

用取与不取来枚举选出m个点的方案，既然m个点选定了，那么分母就确定了，分子通过最小生成树算出最小。

#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;

typedef long long ll;
const int maxn=20;
int n,m,d[maxn],a[maxn][maxn],cnt,father[maxn];

struct edge{
    int u,v,w;
    edge(int u0=0,int v0=0,int w0=0){
        u=u0,v=v0,w=w0;
    }
    friend bool operator <(edge x,edge y){
        return x.w<y.w;
    }
}e[maxn*maxn];

int find(int x){
    if(father[x]!=x){
        father[x]=find(father[x]);
    }
    return father[x];
}

int getAns(vector <int> v){
    for(int i=0;i<=n;i++) father[i]=i;
    cnt=0;
    for(int i=0;i<v.size();i++){
        for(int j=0;j<v.size();j++){
            if(a[v[i]][v[j]]) e[cnt++]=edge(v[i],v[j],a[v[i]][v[j]]);
        }
    }
    sort(e,e+cnt);
    int tmp=0;
    for(int i=0;i<cnt;i++){
        if(find(e[i].u)!=find(e[i].v)){
            father[find(e[i].v)]=find(e[i].u);
            tmp+=e[i].w;
        }
    }
    return tmp;
}

void solve(){
    int ansm=1,ansz=(1<<30);
    vector <int> ansv;
    for(int i=0;i<(1<<n);i++){
        vector <int> v;
        int tmpm=0;
        for(int t=0;t<n;t++){
            if(i&(1<<t)){
                v.push_back(t);
                tmpm+=d[t];
            }
        }
        if(v.size()==m){
            int tmpz=getAns(v);
            if((ll)tmpz*(ll)ansm<(ll)ansz*(ll)tmpm){
                ansz=tmpz;
                ansm=tmpm;
                ansv=v;
            }
        }
    }
    for(int i=0;i<ansv.size();i++){
        if(i>0) printf(" ");
        printf("%d",ansv[i]+1);
    }
    printf("\n");
}

int main(){
    while(scanf("%d%d",&n,&m)!=EOF && (m||n)){
        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",&a[i][j]);
            }
        }
        solve();
    }
    return 0;
}