hdu1054最小顶点覆盖

最小定点覆盖是指这样一种情况：

图G的顶点覆盖是一个顶点集合V，使得G中的每一条边都接触V中的至少一个顶点。我们称集合V覆盖了G的边。最小顶点覆盖是用最少的顶点来覆盖所有的边。顶点覆盖数是最小顶点覆盖的大小。

相应地，图G的边覆盖是一个边集合E，使得G中的每一个顶点都接触E中的至少一条边。如果只说覆盖，则通常是指顶点覆盖，而不是边覆盖。

在二分图中：最大匹配数=最小顶点覆盖数；

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:

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)

Output

1
2

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)题意：找该图的最小顶点覆盖数题解：二分图匹配vector存图，貌似用邻接矩阵会超时（好像还可以用树形dp做，等学了再来更新做法！）

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define pi acos(-1)
#define ll long long
#define mod 1000000007

using namespace std;

const int N=1500+5,maxn=100+5,inf=0x3f3f3f3f;

int color[N];
bool used[N];
vector<int>v[N];

bool match(int x)
{
    for(int i=0;i<v[x].size();i++)
    {
        int t=v[x][i];
        if(!used[t])
        {
            used[t]=1;
            if(color[t]==-1||match(color[t]))
            {
                color[t]=x;
                return 1;
            }
        }
    }
    return 0;
}
int main()
{
    int n;
    while(cin>>n){
        for(int i=0;i<n;i++)v[i].clear();
        for(int i=0;i<n;i++)
        {
            int a,b,k;
            scanf("%d:(%d)",&a,&b);
            while(b--){
                cin>>k;
                v[a].push_back(k);
                v[k].push_back(a);
            }
        }
        int ans=0;
        memset(color,-1,sizeof color);
        for(int i=0;i<n;i++)
        {
            memset(used,0,sizeof used);
            ans+=match(i);
        }
        cout<<ans/2<<endl;
    }
    return 0;
}