hdu 2489(状态压缩+最小生成树)

Minimal Ratio Tree

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3899 Accepted Submission(s): 1196

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

Sample Output

1 3
1 2

题意：在n个点中选m个点出来，求使上面的公式最小的m个点的组合。

题解：点比较少,状压31MS。。先用状态压缩选择m个点出来，然后再对选出来的点进行公式的计算。

#include <stdio.h>
#include <iostream>
#include <string.h>
#include <algorithm>
#include <math.h>
using namespace std;
const int N = 16;
const int INF = 9999999999;
int graph[N][N];
int weight[N];
int state[1<<N];
int a[N],result[N];;
int n,m;
int vis[N],low[N];
void input(){
    for(int i=1;i<=n;i++) scanf("%d",&weight[i]);
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            scanf("%d",&graph[i][j]);
        }
    }
}
bool check(int num){
    int cnt =0;
    while(num){
        if(num&1) cnt++;
        num = num>>1;
    }
    if(cnt==m) return true;
    return false;
}
void init(int &k){
    for(int i=0;i<(1<<n);i++){
        if(check(i)) state[k++]=i;
    }
    k--;
}
int prim(int n,int pos){
    memset(vis,0,sizeof(vis));
    for(int i=0;i<n;i++){
        low[a[i]] = graph[pos][a[i]];
    }
    vis[pos] = true;
    int cost = 0;
    for(int i=1;i<n;i++){
        int Min = INF;
        for(int j=0;j<n;j++){
            if(!vis[a[j]]&&low[a[j]]<Min){
                pos = a[j];
                Min = low[a[j]];
            }
        }
        vis[pos] = true;
        cost +=Min;
        for(int j=0;j<n;j++){
            if(!vis[a[j]]&&low[a[j]]>graph[pos][a[j]]) low[a[j]] = graph[pos][a[j]];
        }
    }
    return cost;
}
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF,n+m){
        int k=0;
        init(k);
        int min_id=N;
        double _ratio = INF*1.0;
        input();

        for(int i=0;i<=k;i++){
            int num = state[i];
            int t =0,V=0;
            for(int j=0;j<n;j++){
                if((num>>j)&1){  ///这里代表第j+1个点要选进去
                    a[t++] = j+1;
                    V += weight[j+1];
                }
            }
            int cost = prim(t,a[0]);
            double temp1 = cost*1.0/V;
            if(temp1<_ratio){
                _ratio = temp1;
                for(int i=0;i<t;i++) result[i] = a[i];
            }
        }
        for(int i=0;i<m-1;i++){
            printf("%d ",result[i]);
        }
        printf("%d
",result[m-1]);
    }
}