(树形DP) zoj 3201
You're given a tree with weights of each node, you need to find the maximum subtree of specified size of this tree.
Tree Definition
A tree is a connected graph which contains no cycles.
Input
There are several test cases in the input.
The first line of each case are two integers N(1 <= N <= 100), K(1 <= K <= N), where N is the number of nodes of this tree, and K is the subtree's size, followed by a line with N nonnegative integers, where the k-th integer indicates the weight of k-th node. The following N - 1 lines describe the tree, each line are two integers which means there is an edge between these two nodes. All indices above are zero-base and it is guaranteed that the description of the tree is correct.
Output
One line with a single integer for each case, which is the total weights of the maximum subtree.
Sample Input
3 1 10 20 30 0 1 0 2 3 2 10 20 30 0 1 0 2
Sample Output
30 40
题意:选K个节点的子树的最大值
dp[i][j]以i为父亲节点的选j个节点的最大值
#include<iostream> #include<cstdio> #include<cstring> #include<string> #include<cmath> #include<cstdlib> #include<algorithm> #include<vector> using namespace std; int n,k,dp[105][105],val[105]; vector<int> e[105]; void dfs(int u,int father) { dp[u][1]=val[u]; for(int i=0;i<e[u].size();i++) { int v=e[u][i]; if(v==father) continue; dfs(v,u); for(int j=k;j>=1;j--) { for(int t=1;t<=j;t++) dp[u][j]=max(dp[u][j],dp[u][t]+dp[v][j-t]); } } } int main() { while(scanf("%d%d",&n,&k)!=EOF) { for(int i=0;i<n;i++) { e[i].clear(); for(int j=0;j<n;j++) dp[i][j]=0; } for(int i=0;i<n;i++) scanf("%d",&val[i]); for(int i=1;i<n;i++) { int x,y; scanf("%d%d",&x,&y); e[x].push_back(y); e[y].push_back(x); } dfs(0,-1); int ans=0; for(int i=0;i<n;i++) ans=max(ans,dp[i][k]); printf("%d\n",ans); } return 0; }