POJ 1463 Strategic game 樹形DP

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

題意：在一棵樹形道路上的結點中佈置士兵，每個士兵可以守護與其結點直接相連的邊，問最少要佈置多少士兵。

分析：每個節點可選擇是否佈置士兵看守，分別用dp[v][0]和dp[v][1]表示該情況下以v爲根節點的子樹最少需要佈置的士兵數量。

初始狀態：dp[v][0]=0;dp[v][1]=1;

非葉節點v：若節點v不佈置士兵，則其所有直接相連的子節點都必須佈置士兵，所以dp[v][0] += Σdp[u][1]（u爲v的各子節點）；

                    若節點v佈置士兵，則其各子樹可儘量放最少的士兵數，dp[v][1] +=Σmin(dp[u][0], dp[u][1])。

最終結果爲min(dp[root][0],dp[root][1]).。

#include<iostream>
#include<cstring>
#include<vector>
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
const int maxn=1505;
int dp[maxn][2];
vector<int> G[maxn]; 

void dfs(int v,int root)
{
	for(int i=0;i<G[v].size();i++)
	{
		int u=G[v][i];
		if(u==root) continue;
		dfs(u,v);
		dp[v][0]+=dp[u][1];//不佈置士兵 
		dp[v][1]+=min(dp[u][0],dp[u][1]);//佈置士兵 
	}
}
int main()
{
	int n;
	while(cin>>n)
	{
		//初始化 
		mem(dp,0);
		for(int i=0;i<n;i++)
			G[i].clear();
		
		for(int i=0;i<n;i++)
		{
			int v,num;
			scanf("%d:(%d)",&v,&num);
			dp[v][1]=1;
			for(int j=0;j<num;j++)
			{
				int u;
				scanf("%d",&u);
				G[v].push_back(u);
				G[u].push_back(v);
			}
		}
		dfs(0,-1);
		cout<<min(dp[0][0],dp[0][1])<<endl;
		
	}
	return 0;
}