hdu-2489 Minimal Ratio Tree（DFS+最小生成树）

题目链接：点击打开链接

Minimal Ratio Tree

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3909    Accepted Submission(s): 1203

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

3 230 20 100 6 26 0 32 3 02 21 10 22 00 0

 

Sample Output

1 31 2

题意：有n个点。接着输入n个点的权值，再输入图，代表了各条边的权值，题目说明了主对角线全为0，因此不用特别处理i==j的情况。

思路：n只有15， 想到用DFS全遍历，当然，裸DFS会TLE，于是只要加点剪枝就可以了。 31msAC。

这里剪枝的条件是1：第一个数+m<=n为什么呢？ 假设n是5，m是3，那么第一个数至多是3吧（3,4,5），大于3就不行了

2：我们可以让序列严格递增，这样可以减少很少遍历，（同样几个数字，不剪枝会改变顺序出现很多次）

于是问题就简单了，题目要求m个点，每次我们遍历出符合条件的m个点时，其实m个点的权值就确定了，那么要让边和/点的权值结果最小，自然就是让边和最小了。

所以此时只要在这m个点求一次最小生成树得到最短长度就能得到一个题目要求的"最小比例树"了，然后再在所有符合条件的结果中取最小的，就是最终答案了~

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;#define N 20#define INF 1<<30#define eps 1e-8int n,m;double minn;int ma[N][N];int p[N],ans[N];int f[N],flag[N];int d[N],flag1[N];int prim(){    int ret=0;///最小生成树大小    memset(flag,0,sizeof(flag));    for(int i=1;i<m;i++)            d[i]=ma[f[0]][f[i]];    flag[0]=1;    for(int i=1;i<m;i++)    {        int v=-1;        for(int j=0;j<m;j++)            if(!flag[j]&&(v==-1||(d[v]>d[j])))                v=j;        if(v==-1)            return -1;        ret+=d[v];        flag[v]=1;        for(int j=0;j<m;j++)            if(!flag[j]&&d[j]>ma[f[v]][f[j]])                    d[j]=ma[f[v]][f[j]];    }    return ret;}void dfs(int k){    if(k==m)    {        double sum=0;        for(int i=0;i<m;i++)            sum+=p[f[i]];        double ret=prim();        if(fabs(ret+1)<eps)            return;        if(minn-(ret/sum)>eps)        {            minn=ret/sum;            for(int i=0;i<m;i++)                ans[i]=f[i];        }        return;    }    for(int i=0;i<n;i++)    {        if(k==0&&i+m>n)             return;        if(k>0&&i<f[k-1])            continue;        if(!flag1[i])        {            flag1[i]=1;            f[k]=i;            dfs(k+1);            flag1[i]=0;        }    }}int main(){    while(~scanf("%d %d",&n,&m)&&n)    {        for(int i=0;i<n;i++)            scanf("%d",&p[i]);        for(int i=0;i<n;i++)            for(int j=0;j<n;j++)            scanf("%d",&ma[i][j]);            memset(flag1,0,sizeof(flag1));            minn=INF;        dfs(0);        for(int i=0;i<m-1;i++)            printf("%d ",ans[i]+1);            printf("%d\n",ans[m-1]+1);    }    return 0;}