Strategic game(POJ 1463 树形DP)

Strategic game

Time Limit: 2000MS		Memory Limit: 10000K
Total Submissions: 7490		Accepted: 3483

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

一道简单的树形DP,状态转移方程：和上一次做的题目基本一样

　　dp[node][0]+=dp[tree[node][i]][1];

　　dp[node][1]+=min(dp[tree[node][i]][0],dp[tree[node][i]][1]); 然后+1

一开始写成 dp[node][1]+=dp[tree[node][i]][0];

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <cstring>
 5 using namespace std;
 6 #define Max 1506
 7 int dp[Max][2],fa[Max],tree[Max][12];
 8 int n;
 9 int num[Max];
10 void tree_dp(int node)
11 {
12     int i;
13     for(i=0;i<num[node];i++)
14     {
15         tree_dp(tree[node][i]);
16         dp[node][0]+=dp[tree[node][i]][1];
17         dp[node][1]+=min(dp[tree[node][i]][0],dp[tree[node][i]][1]);
18     }
19     dp[node][1]++;
20     return;
21 }
22 int main()
23 {
24     int i,j;
25     int a,b;
26     int k,root,f;
27     freopen("in.txt","r",stdin);
28     while(scanf("%d",&n)!=EOF)
29     {
30         memset(dp,0,sizeof(dp));
31         memset(tree,0,sizeof(tree));
32         for(i=0;i<Max;i++)
33             fa[i]=-1;
34         for(i=0;i<n;i++)
35         {
36             scanf("%d%*c%*c%d%*c",&f,&k);
37             num[f]=k;
38             for(j=0;j<k;j++)
39             {
40                 scanf("%d",&tree[f][j]);
41                 fa[tree[f][j]]=f;
42             }
43         }
44         root=0;
45         while(fa[root]!=-1)
46             root=fa[root];
47         tree_dp(root);
48         printf("%d
",min(dp[root][0],dp[root][1]));
49     }
50     return 0;
51 }