UVA 1292 十二 Strategic game

Strategic game

Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu

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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 file 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). 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).

 

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

 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <vector>
 4 #include <algorithm>
 5 using namespace std;
 6 
 7 vector <int> a[1505];
 8 int dp[1505][2],vis[1505];
 9 
10 void dfs(int o)
11 {
12     vis[o]=1;dp[o][0]=0;dp[o][1]=1;
13     for(int i=0;i<a[o].size();i++)
14     {
15         int v=a[o][i];
16         if(vis[v]==1) continue;
17         dfs(v);
18         dp[o][1]=dp[o][1]+min(dp[v][1],dp[v][0]);
19         dp[o][0]=dp[o][0]+dp[v][1];
20     }
21     return;
22 }
23 
24 int main()
25 {
26     int n,m,x,y;
27     int i,j,k;
28     while(scanf("%d",&n)!=EOF)
29     {
30         for(i=0;i<=n;i++)
31             a[i].clear();
32         memset(dp,0,sizeof(dp));
33         memset(vis,0,sizeof(vis));
34         while(n--)
35         {
36             scanf("%d:(%d)",&x,&m);
37             for(i=1;i<=m;i++)
38             {
39                 scanf("%d",&y);
40                 a[x].push_back(y);
41                 a[y].push_back(x);
42             }
43         }
44 
45         dfs(0);
46         printf("%d\n",min(dp[0][1],dp[0][0]));
47     }
48     return 0;
49 }

View Code

 

转载于:https://www.cnblogs.com/cyd308/p/4771597.html