POJ 1463 Strategic game （树形DP）

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

dp表示第i个节点放不放士兵花的最小值。
从叶子节点回溯，最后到根节点，输出答案。
开始dp0初始化为0，表示不放士兵，dp1初始化为1，自然表示放士兵。
核心语句：

dp[root][0] += dp[g[root][i]][1];//父节点不放，子节点一定放
dp[root][1] += min(dp[g[root][i]][0], dp[g[root][i]][1]);//父节点放了，子节点可放可不放，选小的

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <map>
#include <string>
#include <vector>
//#include <iostream>
using namespace std;
//for(i=1;i<n;i++)
//scanf("%d",&n);
//printf("\n",);
int dp[1600][2];
vector<int>g[1600];
int father[1600];

int dfs(int root)
{
    for(int i=0; i<g[root].size(); i++)
    {
         dfs(g[root][i]);
    }

    for(int i=0; i<g[root].size(); i++)
    {
        dp[root][0] += dp[g[root][i]][1];
        dp[root][1] += min(dp[g[root][i]][0], dp[g[root][i]][1]);
    }

    return 0;
}


int main()
{
    int i,j,k,l,n,m;

    while(scanf("%d",&n)!=EOF)
    {

        for(i=0;i<=n;i++)
        {
            g[i].clear();
            dp[i][1]=1;dp[i][0]=0;
            father[i]=-1;
        }


        for(i=1;i<=n;i++)
        {
           scanf("%d:(%d)",&k,&l);

           for(j=0;j<l;j++)
           {
              scanf("%d",&m);
              g[k].push_back(m);
              father[m]=k;
           }
        }

        int root=1;

        while(father[root]!=-1)
        root=father[root];

        dfs(root);

        printf("%d\n",min(dp[root][0],dp[root][1]));

    }

    return 0;
}