POJ 1463

Strategic game

Time Limit: 2000MS

Memory Limit: 10000KB

64bit IO Format: %I64d & %I64u

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


題意：有一棵樹，在一個節點上放一個戰士，相鄰的點也會受到保護，問最少需要多少戰士，纔可以使所有的點都得到保護，即求樹的最小點覆蓋~   樹狀DP,和POJ 2342大同小異~
AC代碼：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
#define maxn 1500+5
int n,dp[maxn][2],fa[maxn],vis[maxn];
void dfs(int x)
{
  vis[x] = 1;
  for(int i = 0;i < n;i ++)
  {
     if(!vis[i] && fa[i] == x)
     {
      dfs(i);
      dp[x][0] += dp[i][1];
      dp[x][1] += min(dp[i][0], dp[i][1]);
     }
  }
  return;
}
int main()
{
    int from, to, root, m;
    char c1, c2;
    while(scanf("%d", &n) != EOF)
    {
     for(int i = 0;i < n;i ++)
     fa[i] = -1;
     memset(vis, 0, sizeof(vis));
     for(int i = 0;i < n;i ++)
     {
        scanf("%d%c%c",&from,&c1,&c2);
        scanf("%d", &m);
        getchar();
        while(m --)
        {
        scanf("%d",&to);
        fa[to] = from;
        }
     }
     for(int i = 0;i < n;i ++)
     {
       if(fa[i] == -1) root = i;
        dp[i][0] = 0;
        dp[i][1] = 1;
     }
     dfs(root);
     printf("%d\n",min(dp[root][0], dp[root][1]));
    }
    return 0;
}