poj 1463 Strategic game（树形dp，最小边覆盖集）

Time Limit: 2000MS		Memory Limit: 10000K
Total Submissions: 7501		Accepted: 3489

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_identifier_{number_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

Source

Southeastern Europe 2000

题目大意：

        有一个树形村庄，节点与道路相连，问在节点上最少放多少个士兵可以使得所有边被士兵观察到，每个士兵能看到与节点邻接的边。

解题思路：

        dp[i][j]表示第i个点是否放置士兵的使得子树边全部被观察到的最小士兵数。dp[i][0]=sigema(dp[j][1]);dp[i][1]=sigema(min(dp[j][0],dp[j][1]),所求的就是min(dp[1][0],dp[1][1]).

#include<stdio.h>
#include<map>
#include<string>
#include<cstring>
#include<algorithm>
#include<vector>
#include<iostream>
#define maxn 1550
#define c(a) memset(a,0,sizeof(a))
#define c_1(a) memset(a,-1,sizeof(a))
using namespace std;
vector<int>cl[maxn];
int  dp[maxn][2];
bool v[maxn];
int dfs(int i, int j)
{
	if (dp[i][j] != -1)return dp[i][j];
	int ans = 0;
	if (j == 1)
	{
		ans++;
		for (int k = 0; k < cl[i].size(); k++)
		{
			int a = dfs(cl[i][k], 0);
			int b = dfs(cl[i][k], 1);
			ans += min(a, b);
		}
		return dp[i][j] = ans;
	}
	else
	{
		for (int k = 0; k < cl[i].size(); k++)
		{
			ans += dfs(cl[i][k], 1);
		}
		return dp[i][j] = ans;
	}
}
int main()
{
	int n;
	while (scanf("%d", &n)==1)
	{
		c_1(dp); c(v);
		int a,b,s;
		for (int i = 0; i<n; i++)cl[i].clear();
		for (int i = 0; i < n; i++)
		{
			scanf("%d:(%d)",&s ,&a);
			for (int j = 0; j < a; j++)
			{
				scanf("%d", &b);
				cl[s].push_back(b);
				v[b] = 1;
			}
		}
		for (int i = 0; i < n; i++)if (cl[i].size() == 0) { dp[i][0] = 0; dp[i][1] = 1; }
		int x=0;
		for (int i = 0; i < n; i++)if (!v[i]) x = x+min(dfs(i, 1), dfs(i, 0)); 
		printf("%d\n",x);
	}
}