浙江财经大学
信工学院ACM集训队

HDU 2489 Minimal Ratio Tree(DFS+Prime)

本文由 Ocrosoft 于 2016-07-30 10:17:12 发表

Minimal Ratio Tree

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4009    Accepted Submission(s): 1241

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

Sample Output
	
1 3 1 2
 

Solution

题意:给出一个n个节点的图,每个结点都有n-1条出边,每条边每个节点都有权值,要求从中选出m个点,使得边权值和与结点权值和的商最小,最后升序输出所选择的那些点。如果商相同,字典序升序输出。

由于最多15个结点,可以枚举m个点的情况。对所有情况进行prime。

没有什么精度问题,讨论版的是错的。一开始WA了将近10次,最后才发现,错在dfs函数,因为字典序的问题,需要把选择这个点写在不选择这个点的前面。

#include <set>
#include <map>
#include <list>
#include <cmath>
#include <stack>
#include <queue>
#include <ctime>
#include <string>
#include <cstdio>
#include <vector>
#include <cctype>
#include <climits>
#include <sstream>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#define ms(a) memset(a,0,sizeof(a))
#define INF INT_MAX
#define MAXN 10+10
using namespace std;
int n,m;
int mp[MAXN][MAXN],lowcast[MAXN],nodecast[MAXN];
bool vis[MAXN],use[MAXN],tans[MAXN];
double res;
int prime(int u0)
{
    int ans=0;
    int pos=u0;//pos记录为起点
    memcpy(vis,use,sizeof(use)),vis[u0]=0;//清空vis数组,标记起点
    for(int i=1;i<=n;i++)vis[i]=!vis[i];
    for(int i=1; i<=n; i++)//初始化lowcast数组
        lowcast[i]=(i==pos?0:mp[pos][i]);
    for(int i=1; i<m; i++)
    {
        int minn=INF;
        for(int j=1; j<=n; j++) //找到没访问过的而且最近的邻近点
            if(!vis[j]&&minn>lowcast[j])minn=lowcast[j],pos=j;
        if(minn==INF)return 0;
        ans+=minn;
        vis[pos]=1;//标记这个邻近点
        for(int j=1; j<=n; j++) //更新lowcast数组
            if(!vis[j]&&lowcast[j]>mp[pos][j])lowcast[j]=mp[pos][j];
    }
    return ans;
}
void dfs(int cur,int cnt)
{
    if(cnt==m)
    {
        double a,b=0,c;
        for(int i=1; i<=n; i++)
        {
            if(use[i])
            {
                a=prime(i);
                break;
            }
        }
        for(int i=1; i<=n; i++)
            if(use[i])b+=nodecast[i];
        c=a/b;
        if(c<res)res=c,memcpy(tans,use,sizeof(use));
        return;
    }
    if(cur>n)return;
    if(cnt>m)return;
    use[cur]=1;
    dfs(cur+1,cnt+1);
    use[cur]=0;
    dfs(cur+1,cnt);
}
int main()
{
    while(cin>>n>>m&&(n||m))
    {
        for(int i=1; i<=n; i++)
            scanf("%d",&nodecast[i]);
        for(int i=1; i<=n; i++)
            for(int j=1; j<=n; j++)
                scanf("%d",&mp[i][j]);
        res=INF;
        dfs(1,0);
        bool flag=1;
        for(int i=1;i<=n;i++)
        {
            if(tans[i])
            {
                if(flag)printf("%d",i),flag=0;
                else printf(" %d",i);
            }
        }
        printf("\n");
    }
    return 0;
}

 

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