POJ 1947 Rebuilding Roads(树形DP)

题目链接

树形DP慢慢来，慢慢学习。这个题，就是树上的背包。

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <iostream>
 4 #include <cmath>
 5 #include <algorithm>
 6 using namespace std;
 7 #define INF 100000000
 8 int dp[501][501];
 9 int first[501];
10 int o[501],m,t;
11 struct node
12 {
13     int u,v,next;
14 }edge[1000000];
15 void CL()
16 {
17     t = 1;
18     memset(first,-1,sizeof(first));
19     memset(o,0,sizeof(o));
20 }
21 void add(int u,int v)
22 {
23     edge[t].u = u;
24     edge[t].v = v;
25     edge[t].next = first[u];
26     first[u] = t++;
27 }
28 void dfs(int x)
29 {
30     int i,j,k,temp;
31     for(i = 0;i <= m;i ++)
32     dp[x][i] = INF;
33     dp[x][1] = 0;
34     for(i = first[x];i != -1;i = edge[i].next)
35     {
36         dfs(edge[i].v);
37         for(j = m;j >= 1;j --)
38         {
39             temp = dp[x][j] + 1;
40             for(k = 1;k < j;k ++)
41             {
42                 temp = min(temp,dp[edge[i].v][j-k]+dp[x][k]);
43             }
44             dp[x][j] = temp;
45         }
46     }
47 }
48 int main()
49 {
50     int n,i,u,v,ans;
51     while(scanf("%d%d",&n,&m)!=EOF)
52     {
53         CL();
54         for(i = 1;i < n;i ++)
55         {
56             scanf("%d%d",&u,&v);
57             add(u,v);
58             o[v] = 1;
59         }
60         for(i = 1;i <= n;i ++)
61         {
62             if(o[i] == 0)
63             {
64                 dfs(i);
65                 ans = dp[i][m];
66                 break;
67             }
68         }
69         for(i = 1;i <= n;i ++)
70         {
71             ans = min(ans,dp[i][m]+1);
72         }
73         printf("%d\n",ans);
74     }
75     return  0;
76 }

 

 

 

posted @ 2013-07-20 10:36  Naix_x  阅读(186)  评论(0)  编辑  收藏  举报