HDU 2489 minimal ratio tree (Data Structure-minimal spanning tree)

Minimal ratio tree
Problem descriptionfor 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.
 
Inputinput 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 discounted rical 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 (blocks t for the ones on the diagonal of the matrix) are integers and in the range of [1,100].

The figure below tables strates the first test case in sample input. Node 1 and node 3 form the minimal ratio tree.

 
Outputfor each test case output one line contains a sequence of the M nodes which constructs the minimal ratio tree. nodes shoshould 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

 
Source2008 Asia Regional Beijing
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Question:

Give you a picture of N points, each point has the right value, ask you to select M points, so that the minimum, the output scheme.

Solution:

The scheme of selecting M vertices by enumeration with and without. Since m vertices are selected, the denominator is determined, and the numerator calculates the minimum through the minimal spanning tree.

Solution code:

#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;}