I - Strategic Game （二分圖最大匹配的性質）

I - Strategic Game

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)

Sample Output

1
2

題意：

         Bob想保衛他的城鎮，所以就安排士兵巡邏。因爲兵力有限，就儘可能少的安排士兵，並使士兵能儘快到達鄰近點。

思路：

         二分圖最大匹配數 等於 這個圖的最小點覆蓋數

最小點覆蓋——假如選了一個點就相當於 覆蓋了以它爲端點所有邊，則需要選擇最少的點圖中覆蓋所有邊。（本題關鍵）

     

注意：

       本代碼建立的是 無向圖 ，所以 最大匹配 / 2 ；

代碼：

#include<map>
#include<queue>
#include<math.h>
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define MM 2007
int line[MM][MM],book[MM],man[MM];
int n,m;
bool Find(int x)
{
    for(int i=0; i<n; i++)
    {
        if(line[x][i] && !book[i])
        {
            book[i]=1;
            if(!man[i] || Find(man[i]))
            {
                man[i]=x;
                return true;
            }
        }
    }
    return false;
}
int match()
{
    int ans=0;
    for(int i=0; i<n; i++)
    {
        mem(book,0);
        if(Find(i))
            ans++;
    }
    return ans/2;
}
int main()
{
    while(~scanf("%d",&n))
    {
        mem(line,0);
        mem(man,0);
        int jd,g,x;
        for(int i=1; i<=n; i++)
        {
            scanf("%d:(%d)",&jd,&g);
            for(int j=1;j<=g;j++)
            {
                scanf("%d",&x);
                line[jd][x]=line[x][jd]=1;  //建立無向圖
            }
        }
        printf("%d\n",match());
    }
    return 0;
}