本文共 2119 字，大约阅读时间需要 7 分钟。

题目：

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:

Input

40:(1) 11:(2) 2 32:(0)3:(0)53:(3) 1 4 21:(1) 02:(0)0:(0)4:(0)

Output

12

Sample Input

40:(1) 11:(2) 2 32:(0)3:(0)53:(3) 1 4 21:(1) 02:(0)0:(0)4:(0)

Sample Output

12

思路：

这题感觉跟二分图的最大匹配数很像，因为二分图所组成的组数中的点可以通过一个点来统一管理。

套用了匈牙利算法，A了。

代码如下：

#include 
   
    #include 
    
     #include 
     
      #include 
      
       using namespace std;const int maxn=1505;int head[maxn];int n;int match[maxn];int vis[maxn];struct edge{    int to;    int next;}edge[maxn<<1];void  add (int id,int u,int v){    edge[id].to=v;    edge[id].next=head[u];    head[u]=id;}bool Find(int x){    for (int i=head[x];i!=-1;i=edge[i].next)    {        int u=edge[i].to;        if(!vis[u])        {            vis[u]=1;            if(match[u]==-1||Find(match[u]))            {                match[u]=x;                return true;            }        }    }    return false;}int algor(){    int sum=0;    memset (match,-1,sizeof(match));    for (int i=0;i
      
     
    
   

 

转载地址：http://zzoen.baihongyu.com/