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:

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)

1
2

最小顶点覆盖，一直超时，后来才意识到图是一个稀疏图，用矩阵存图果断超时。用邻接表轻松AC！

#include"stdio.h"
#include"string.h"
#include"queue"
#include"vector"
using namespace std;
#define N 1505
#define max(a,b) (a>b?a:b)
#define min(a,b) (a<b?a:b)
const int inf=0x7fffffff;
int mark[N],n,t;
int link[N],head[N];
struct node
{
    int u,v,next;
}g[N*10];
void add(int u,int v)
{
    g[t].u=u;
    g[t].v=v;
    g[t].next=head[u];
    head[u]=t++;
}
int find(int u)
{
    int i,v;
    for(i=head[u];i!=-1;i=g[i].next)
    {
        v=g[i].v;
        if(!mark[v])
        {
            mark[v]=1;
            if(link[v]==-1||find(link[v]))
            {
                link[v]=u;
                return 1;
            }
        }
    }
    return 0;
}
int main()
{
    int i,u,q,v;
    while(scanf("%d",&n)!=-1)
    {
        memset(head,-1,sizeof(head));
        memset(link,-1,sizeof(link));
        t=0;
        for(i=0;i<n;i++)
        {
            scanf("%d:(%d)",&u,&q);
            while(q--)
            {
                scanf("%d",&v);
                add(u,v);
                add(v,u);  //建立的双向边，所以答案最后除以二
            }
        }
        int ans=0;
        for(i=0;i<n;i++)
        {
            memset(mark,0,sizeof(mark));
            ans+=find(i);
        }
        printf("%d\n",ans/2);
    }
    return 0;
}