HDU - 1054 Strategic Game 树形DP

Posted qldabiaoge

tags:

篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了HDU - 1054 Strategic Game 树形DP相关的知识,希望对你有一定的参考价值。

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: 
技术分享图片

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
 1 #include <cstdio>
 2 #include <algorithm>
 3 #include <cstring>
 4 using namespace std;
 5 const int maxn = 1e5 + 10;
 6 const int INF = 0x7fffffff;
 7 int n, dp[maxn][2], head[maxn], tot;
 8 struct node {
 9     int v, next;
10 } edge[maxn];
11 void init() {
12     tot = 0;
13     memset(head, -1, sizeof(head));
14 }
15 void add(int u, int v) {
16     edge[tot].v = v;
17     edge[tot].next = head[u];
18     head[u] = tot++;
19     edge[tot].v = u;
20     edge[tot].next = head[v];
21     head[v] = tot++;
22 }
23 void solve(int x, int fa) {
24     dp[x][0] = 0;
25     dp[x][1] = 1;
26     for (int i = head[x] ; i != -1 ; i = edge[i].next) {
27         int v = edge[i].v;
28         if (v == fa) continue;
29         solve(v, x);
30         dp[x][1] += min(dp[v][0], dp[v][1]);
31         dp[x][0] += dp[v][1];
32     }
33 }
34 int main() {
35     while(scanf("%d", &n) != EOF) {
36         init();
37         int x, y, z;
38         for (int i = 0 ; i < n ; i++) {
39             scanf("%d:(%d)", &x, &y);
40             while(y--) {
41                 scanf("%d", &z);
42                 add(x, z);
43             }
44         }
45         solve(0, -1);
46         printf("%d
", min(dp[0][0], dp[0][1]));
47     }
48     return 0;
49 }

 













以上是关于HDU - 1054 Strategic Game 树形DP的主要内容,如果未能解决你的问题,请参考以下文章

HDU - 1054 Strategic Game 树形DP

HDU 1054 Strategic Game(树形dp)

HDU——T 1054 Strategic Game

HDU - 1054 Strategic Game

HDU1054 Strategic Game——匈牙利算法

HDU 1054 Strategic Game