HDU 1054 Strategic Game 最小顶点覆盖

Strategic Game

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5068    Accepted Submission(s): 2321

Problem Description

Bob enjoys playing computer games, especially strategic games, but sometimes he cannot find the solution fast enough and then he is very sad. Now he has the following problem. He must defend a medieval city, the roads of which form a tree. He has to put the minimum number of soldiers on the nodes so that they can observe all the edges. Can you help him?

Your program should find the minimum number of soldiers that Bob has to put for a given tree.

The input file contains several data sets in text format. Each data set represents a tree with the following description:

the number of nodes

the description of each node in the following format

node_identifier:(number_of_roads) node_identifier1 node_identifier2 ... node_identifier

or

node_identifier:(0)

The node identifiers are integer numbers between 0 and n-1, for n nodes (0 < n <= 1500). Every edge appears only once in the input data.

For example for the tree:

the solution is one soldier ( at the node 1).

The output should be printed on the standard output. For each given input data set, print one integer number in a single line that gives the result (the minimum number of soldiers). An example is given in the following table:

Sample Input

4
0:(1) 1
1:(2) 2 3
2:(0)
3:(0)
5
3:(3) 1 4 2
1:(1) 0
2:(0)
0:(0)
4:(0)

Sample Output

1 2

/*
HDOJ 1054 最小顶点覆盖
最小顶点覆盖：选出最少的点，这些点的关联的边都被覆盖、
STL中的vector建立图的邻接表存储小顶点覆盖

最小顶点覆盖！最小顶点覆盖 == 最大匹配（双向图）/2；
*/
#include<iostream>
#include<stdio.h>
#include<vector>
using namespace std;
#define N 1501

vector<int> map[N];
int n,f[N],pre[N];

int find(int cur)
{
    int i,k;
    for(i=0;i<map[cur].size();i++)
    {
        k=map[cur][i];
        if(!f[k])
        {
            f[k]=1;
            if(pre[k]==-1||find(pre[k]))
            {
                pre[k]=cur;
                return 1;
            }
        }
    }
    return 0;
}

int main()
{
    int i,u,k,b,sum;
    
    //freopen("test.txt","r",stdin);
    while(scanf("%d",&n)!=EOF)
    {
        memset(pre,-1,sizeof(pre));
        for(i=0;i<n;i++)
            map[i].clear();
            
        for(i=0;i<n;i++)
        {
            scanf("%d:(%d)",&u,&k);
            while(k--)
            {
                scanf("%d",&b);
                map[u].push_back(b);
                map[b].push_back(u);//建双向图 
            }
        }
        
        sum=0;
        for(i=0;i<n;i++)
        {
            memset(f,0,sizeof(f));
            sum+=find(i);
        }
        printf("%d\n",sum/2);
    }
    return 0;
}