ACdream 1032 Component

Component

Time Limit: 5000ms

Memory Limit: 64000KB

This problem will be judged on ACdream. Original ID: 1032
64-bit integer IO format: %lld      Java class name: (No Java Yet)

Given a tree with weight assigned to nodes, find out minimum total weight connected component with fixed number of node.

Input

The first line contains a single integer n.

The second line contains n integers $w_1,w_2,…,w_n. w_i$ denotes the weight of the node i.

The following (n−1) lines with two integers ai and bi, which denote the edge between ai and bi.

Note that the nodes are labled by $1,2,…,n.$

$(1\leq n\leq 2⋅10^3,1\leq w_i\leq 10^5)$

Output

$n$ integers $c_1,c_2,…,c_n. c_i$ stands for the minimum total weight component with i nodes.

Sample Input

3
1 2 3
1 2
2 3

Sample Output

1 3 6

Source

ftiasch

解题：树形dp

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int maxn = 2010;
 4 vector<int>g[maxn];
 5 int val[maxn],dp[maxn][maxn],n,son[maxn],ans[maxn];
 6 void dfs(int u,int fa){
 7     dp[u][1] = val[u];
 8     dp[u][0] = 0;
 9     son[u] = 1;
10     for(int i = g[u].size()-1; i >= 0; --i){
11         if(g[u][i] == fa) continue;
12         dfs(g[u][i],u);
13         son[u] += son[g[u][i]];
14         for(int j = son[u]; j > 0; --j){
15             for(int k = 1; k <= j; ++k)
16                 dp[u][j] = min(dp[u][j],dp[g[u][i]][j - k] + dp[u][k]);
17         }
18     }
19     for(int i = son[u]; i >= 0; --i)
20         ans[i] = min(ans[i],dp[u][i]);
21 }
22 int main(){
23     while(~scanf("%d",&n)){
24         for(int i = 0; i < maxn; ++i) g[i].clear();
25         for(int i = 1; i <= n; ++i) scanf("%d",val + i);
26         memset(dp,0x3f,sizeof dp);
27         memset(ans,0x3f,sizeof ans);
28         for(int i = 1,u,v; i < n; ++i){
29             scanf("%d%d",&u,&v);
30             g[u].push_back(v);
31             g[v].push_back(u);
32         }
33         dfs(1,-1);
34         for(int i = 1; i <= n; ++i)
35             printf("%d%c",ans[i],i == n?‘\n‘:‘ ‘);
36     }
37     return 0;
38 }