首页 > ACM题库 > HDU-杭电 > hdu 2489 Minimal Ratio Tree-状态压缩-[解题报告]C++
2014
01-26

hdu 2489 Minimal Ratio Tree-状态压缩-[解题报告]C++

Minimal Ratio Tree

问题描述 :

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.

输入:

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.

输出:

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.

样例输入:

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

样例输出:

1 3
1 2

思路:

直接状态压缩暴力枚举

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#define Maxn 30
#define inf 10000000
using namespace std;
int ans,n,m;
int c,d;
int map[Maxn][Maxn];
int lis[Maxn],node[Maxn];
int val[1<<17],sum[1<<17];
int main()
{
    int i,j;
    while(scanf("%d%d",&n,&m),n||m){
        memset(val,48,sizeof(val));
        memset(sum,0,sizeof(sum));
        c=1000000,d=1;
        for(i=1;i<=n;i++)
            scanf("%d",&node[i]);
        memset(map,0,sizeof(map));
        for(i=1;i<=n;i++){
            for(j=1;j<=n;j++){
                scanf("%d",&map[i][j]);
            }
        }
        int N=1<<n;
        N-=1;
        int cnt=0;
        val[0]=0;
        for(i=0;i<=n;i++)
            val[1<<i]=0;
        for(i=1;i<=N;i++){
            cnt=0;
            for(j=0;j<n;j++){
                if(i&(1<<j)) lis[++cnt]=j+1,sum[i]+=node[j+1];
            }
            if(cnt>m) continue;
            if(cnt==m)
            {
                int a=val[i],b=sum[i];
                if(a*d<c*b){
                    ans=i;
                    c=a;
                    d=b;
                }
            }
            int temp;
            for(j=0;j<n;j++){
                if(((1<<j)&i)==0){
                    temp=inf;
                    for(int k=1;k<=cnt;k++){
                        temp=min(temp,map[j+1][lis[k]]);
                    }
                    val[i|(1<<j)]=min(val[i]+temp,val[i|(1<<j)]);
                }
            }
        }
        int num=1;
        for(j=0;j<n;j++)
            if((1<<j)&ans){
                printf("%d",j+1);
                if(num!=m)
                    printf(" ");
                num++;
            }
        printf("\n");
    }
    return 0;
}

 

解题转自:http://www.cnblogs.com/wangfang20/archive/2013/08/25/3280920.html


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    This is really nice to know. I hope it will be successful in the future. Good job on this and keep up the good work.

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