HDU——1054 Strategic Game

　　　　　　　　　　　　Strategic Game

　　　　　　　　　　　　Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
　　　　　　　　　　　　　　　　　　Total Submission(s): 8504    Accepted Submission(s): 4094

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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问题描述
鲍勃喜欢玩电脑游戏，特别是战略游戏，但有时候他无法快速找到解决方案，那么他很伤心。现在他有以下问题。他必须捍卫一座中世纪城市，其道路形成一棵树。他必须将最少数量的士兵放在节点上，以便他们可以观察所有的边缘。你的程序应该找到Bob给给定树的最小兵数。

思路：用最少的节点覆盖所有的边。最小点覆盖裸题=最大匹配数

这个题的数据比较大，用邻接表的话会T，用邻街链表的话可以A

代码：

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define N 1500+10
using namespace std;
bool vis[N];
int n,m,k,x,y,tot,head[N],girl[N],map[N][N];
int read()
{
    ,f=; char ch=getchar();
    ; ch=getchar();}
    +ch-'; ch=getchar();}
    return x*f;
}
struct Edge
{
    int from,to,next;
}edge[N];
int add(int x,int y)
{
    tot++;
    edge[tot].to=y;
    edge[tot].next=head[x];
    head[x]=tot;
}
int find(int x)
{
    for(int i=head[x];i;i=edge[i].next)
    {
        int t=edge[i].to;
        if(!vis[t])
        {
            vis[t]=true;
            ||find(girl[t])) {girl[t]=x;;}
        }
    }
    ;
}
int main()
{
    ,sum,ans;
    while(scanf("%d",&n)!=EOF)
    {
        ans=,sum=;tot=;
        memset(map,,sizeof(map));
        memset(head,,sizeof(head));
        ;i<=n;i++)
        {
            x=read();m=read();
            ,y+),add(y+,x+);}
        }
        memset(girl,-,sizeof(girl));
        ;i<=n;i++)
        {
            memset(vis,,sizeof(vis));
            if(find(i)) ans++;
        }
        printf();
    }
    ;
}