Hdu1054 Strategic Game(最小覆盖点集)

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本文介绍了一种解决最小覆盖点集问题的方法，通过将问题转化为最大二分匹配问题进行求解。具体实现中使用了匈牙利算法，并提供了一个C++代码示例，展示了如何读取树形结构的输入数据并计算所需士兵的最小数量。

Strategic Game

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

Source

Southeastern Europe 2000

————————————————————————————————

题目的意思是给出一棵树，在顶点上放置士兵，使得每条边至少有一个顶点有士兵

思路：即求最小覆盖点集，根据最小覆盖点集=最大二分匹配，求出最大二分匹配即可

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <vector>
#include <set>
#include <stack>
#include <map>
#include <climits>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

const int MAXN=2000;
int uN,vN;  //u,v数目
vector<int>g[MAXN];
int linker[MAXN];
bool used[MAXN];


bool dfs(int u)
{
    int v;
    int siz=g[u].size();
    for(v=0; v<siz; v++)
        if(!used[g[u][v]])
        {
            used[g[u][v]]=true;
            if(linker[g[u][v]]==-1||dfs(linker[g[u][v]]))
            {
                linker[g[u][v]]=u;
                return true;
            }
        }
    return false;
}
int hungary()
{
    int res=0;
    int u;
    memset(linker,-1,sizeof(linker));
    for(u=0; u<uN; u++)
    {
        memset(used,0,sizeof(used));
        if(dfs(u))  res++;
    }
    return res;
}

int main()
{
   int n,m,x,y;
   while(~scanf("%d",&n))
   {
       for(int i=0;i<MAXN;i++)
        g[i].clear();
       for(int i=0;i<n;i++)
       {
           scanf("%d:(%d)",&x,&m);
           for(int j=0;j<m;j++)
           {
              scanf("%d",&y);
              g[x].push_back(y);
              g[y].push_back(x);
           }
       }
       uN=vN=n;
       printf("%d\n",hungary()/2);

   }
    return 0;
}