Strategic Game（二分匹配）_bob enjoys playing computer games, especially stra-优快云博客

本文链接：https://blog.youkuaiyun.com/Enjoying_Science/article/details/40738591

此博客讨论了在一个形成树状结构的城市中，如何最小化士兵数量来确保所有道路都受到监视的问题。通过输入文件描述的树形结构，解决战略游戏中的士兵部署优化挑战。

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Link：http://acm.hdu.edu.cn/showproblem.php?pid=1054

Strategic Game

Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5142 Accepted Submission(s): 2364

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

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My code：

#include<iostream>
#include<algorithm>
#include<vector>
#include<cstring>
#include<cmath>
using namespace std;
const int maxn=1505;
vector<int>map[maxn];
bool vis[maxn];
int match[maxn],ans,un;
bool dfs(int u)
{
for(int i=0;i<map[u].size();i++)
{
if(!vis[map[u][i]])
{
vis[map[u][i]]=true;
if(match[map[u][i]]==-1||dfs(match[map[u][i]]))
{
match[map[u][i]]=u;
return true;
}
}
}
return false;
}
void hangry()
{
ans=0;
memset(match,-1,sizeof(match));
for(int i=0;i<un;i++)
{
memset(vis,false,sizeof(vis));
if(dfs(i))
ans++;
}
ans/=2;
}
int main()
{
int i,u,v,n;
while(scanf("%d",&un)==1)
{
for(i=0;i<maxn;i++)
map[i].clear();
for(i=0;i<un;i++)
{
scanf("%d:(%d)",&u,&n);
while(n--)
{
scanf("%d",&v);
map[u].push_back(v);
map[v].push_back(u);
}
}
hangry();
printf("%d\n",ans);
}
return 0;
}