Strategic game（POJ-1463）

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      

题意：给你一棵树，树上有子节点，问你最少需要多少个看守者可以看到所有的子节点

思路：dp方程的话，就是设一个dp[1000][2]，当取本身节点的时候，就是dp[1000][1]，取子节点的时候就是dp[1000][0];

剩下的话看代码就好了，树形DP入门题。

AC代码：

#include <stdio.h>
#include <string.h>
#include <string>
#include <algorithm>
#include <iostream>
#include <vector>
#include <math.h>
const int maxx=10010;
using namespace std;
int vis[maxx],dp[maxx][2];
vector<int>son[maxx];
void dfs(int u)
{
    dp[u][1]=1;
    dp[u][0]=0;
    for(int i=0; i<son[u].size(); i++)
    {
        dfs(son[u][i]);
        dp[u][0]+=dp[son[u][i]][1];//当前节点放士兵的话，儿子节点就不放士兵
        dp[u][1]+=min(dp[son[u][i]][1],dp[son[u][i]][0]);//当前节点不放士兵的话，儿子节点可放可不放士兵
    }
}
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        memset(vis,-1,sizeof(vis));
        int fa,ch;
        for(int i=0; i<n; i++)
        {
            scanf("%d:(%d)",&fa,&ch);
            son[fa].clear();
            int cnt;
            for(int i=0; i<ch; i++)
            {
                scanf("%d",&cnt);
                son[fa].push_back(cnt);
                vis[cnt]=fa;
            }
        }
        int ans=0;//找根节点
        while(vis[ans]!=-1)
            ans=vis[ans];
        dfs(ans);
        printf("%d\n",min(dp[ans][1],dp[ans][0]));
    }
    return 0;
}