树状dp--B

trategic game
Time Limit:2000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64u
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Status
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.

For example for the tree:


the solution is one soldier ( at the node 1).
Input
The input 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_identifiernumber_of_roads
or
node_identifier:(0)

The node identifiers are integer numbers between 0 and n-1, for n nodes (0 < n <= 1500);the number_of_roads in each line of input will no more than 10. Every edge appears only once in the input data.
Output
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:
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

<span style="color:#3366ff;">/*
***********************************************************************
           author    :     Grant Yuan
           time      :     2014.7.22
           algorithm :     树状dp
           explain   ：    状态转移方程：
                             f[x][1] =1 + Σ min(f[i][0],f[i][1])；// x上放置的士兵，于是它的儿子们可放可不放！
                             f[x][0] = Σ f[i][1]；//x上不放置的士兵，它的儿子们都必须放！
                             (i是x的儿子！！)
                             结果为min(f[root][0], f[root][1])
 ************************************************************************
 */
   
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<vector>
#define MAX 1501

using namespace std;

vector<int> G[MAX];
int dp[MAX][2],n;
int a,b;
int root;

void dfs(int rt)
{
   int len;
   len=G[rt].size();
   for(int i=0;i<len;i++)
   {
       dfs(G[rt][i]);
   }

   for(int i=0;i<len;i++)
   {
      dp[rt][1]+=min(dp[G[rt][i]][0],dp[G[rt][i]][1]);
      dp[rt][0]+=dp[G[rt][i]][1];
   }
   dp[rt][1]+=1;
}

int main()
{
    while(~scanf("%d",&n)){
        memset(dp,0,sizeof(dp));
        memset(G,0,sizeof(G));
        for(int i=0;i<n;i++)
        { 
            scanf("%d:(%d)",&a,&b);
            if(i==0) root=a;
            for(int j=0;j<b;j++)
            {  int t;
               scanf("%d",&t);
                G[a].push_back(t);
            }
        }
        dfs(root);
        int ans;
        ans=min(dp[root][0],dp[root][1]);
        printf("%d
",ans);}
        return 0;
}
</span>