Strategic Game
Time Limit: 20000/10000 MS
(Java/Others) Memory Limit: 65536/32768 K
(Java/Others)
Total Submission(s): 4697 Accepted
Submission(s): 2125
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
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题意：求最少要多少个士兵站岗可使全部点都能观察得到。
因为是树形结构，每个点都会有一条边，最大匹配数即为题解(不证明了..)。
和 hdu1068
差不多，一样的模板，不过这题是求最大匹配数，匈牙利后结果除二即为解。
1 //234MS 280K 1166 B C++
2 #include<stdio.h>
3 #include<string.h>
4 #define N 1505
5 struct node{
6 int v;
7 int next;
8 }edge[10*N];
9 int match[N];
10 int vis[N];
11 int head[N];
12 int n,edgenum;
13 void addedge(int u,int v)
14 {
15 edge[edgenum].v=v;
16 edge[edgenum].next=head[u];
17 head[u]=edgenum++;
18 }
19 int dfs(int x)
20 {
21 for(int i=head[x];i!=-1;i=edge[i].next){
22 int v=edge[i].v;
23 if(!vis[v]){
24 vis[v]=1;
25 if(match[v]==-1 || dfs(match[v])){
26 match[v]=x;
27 return 1;
28 }
29 }
30 }
31 return 0;
32 }
33 int hungary()
34 {
35 int ret=0;
36 memset(match,-1,sizeof(match));
37 for(int i=0;i<n;i++){
38 memset(vis,0,sizeof(vis));
39 ret+=dfs(i);
40 }
41 return ret;
42 }
43 int main(void)
44 {
45 int a,b,m;
46 while(scanf("%d",&n)!=EOF)
47 {
48 edgenum=0;
49 memset(head,-1,sizeof(head));
50 for(int i=0;i<n;i++){
51 scanf("%d:(%d)",&a,&m);
52 while(m--){
53 scanf("%d",&b);
54 addedge(a,b);
55 addedge(b,a);
56 }
57 }
58 printf("%d\n",hungary()/2);
59 }
60 return 0;
61 }
hdu 1054 Strategic Game (二分匹配)