学步园

链接：

HDU ： http://acm.hdu.edu.cn/showproblem.php?pid=2489

POJ  ： http://poj.org/problem?id=3925

题目：

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

题目大意：

有一个n个点的图， 然后给出n*n的邻接矩阵图， 要求这个图的m个结点的子图，使得这个子图所有边之和与所有点之和的商值最小。

分析与总结：

直接dfs枚举出n个点所有的m个点的组合，然后对m个点求最小生成树，便可得出答案。

dfs枚举n个点的m个点组合，对于每个点，要么属于这个组合，要么是不属于，所以复杂度为2^n,  n最大为15， 再加上减枝， 时间足足矣。

代码：

#include<cstdio>
#include<cstring>
#define N 20
int n,m,vis[N], ans[N], pre[N], hash[N];
double G[N][N], weight[N], minCost[N], minRatio;


double prim(){
    memset(hash, 0, sizeof(hash));
    int u;
    for(int i=1; i<=n; ++i)if(vis[i]){
        u=i; break;
    }
    hash[u] = 1;
    double weightSum=0, edgeSum=0;
    for(int i=1; i<=n; ++i)if(vis[i]){
        minCost[i] = G[u][i]; pre[i] = u;
        weightSum += weight[i];
    }

    for(int i=1; i<m; ++i){
        u=-1;
        for(int j=1; j<=n; ++j)if(vis[j]&&!hash[j]){
            if(u==-1 || minCost[u]>minCost[j])
                u=j;
        }
        edgeSum += G[pre[u]][u];
        hash[u] = 1;
        for(int j=1; j<=n; ++j)if(vis[j]&&!hash[j]){
            if(minCost[j] > G[u][j]){
                minCost[j] = G[u][j];
                pre[j] = u;
            }
        }
    }
   return edgeSum/weightSum;
}


void dfs(int u, int num){
    if(num>m) return; 
    if(u==n+1){
        if(num!=m) return;
        double t=prim();
        if(t<minRatio){
            minRatio = t;
            memcpy(ans, vis, sizeof(vis));
        }
        return;
    }
    vis[u] = 1;
    dfs(u+1, num+1);
    vis[u] = 0;
    dfs(u+1, num);
}

int main(){
    while(~scanf("%d%d",&n,&m)){
        if(!n&&!m) break; 
        for(int i=1; i<=n; ++i) 
            scanf("%lf",&weight[i]);
        for(int i=1; i<=n; ++i)
            for(int j=1; j<=n; ++j)
                scanf("%lf",&G[i][j]);
        memset(vis, 0, sizeof(vis));
        minRatio = 100000000;
        dfs(1, 0);
        bool flag=false;
        for(int i=1; i<=n; ++i)if(ans[i]){
            if(flag) printf(" %d", i);
            else { printf("%d",i); flag=true; }
        }
        puts("");
    }
    return 0;

——  生命的意义，在于赋予它意义。
          
     原创 http://blog.csdn.net/shuangde800 ， By   D_Double  (转载请标明)