树形DP专辑-ZOJ3201(Tree of Tree)

ZOJ Problem Set - 3201

Tree of Tree

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Time Limit: 1 Second      Memory Limit: 32768 KB

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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

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Author: LIU, Yaoting
Source: ZOJ Monthly, May 2009

/**************************************************** problem:ZOJ 3201 Tree of Tree* algorithm：树状DP+枚举或者说树状DP+背包* author：sgx* date：2013/09/12****************************************************/#include <iostream>#include <cstdlib>#include <cstring>#include <cstdio>#include <algorithm>using namespace std;const int maxn=110;struct Edge{    int to;    int next;}edge[maxn<<1];int dp[maxn][maxn];int head[maxn],w[maxn];int cnt;int res,m,n;inline int max(int a,int b){    return a>b?a:b;}inline int min(int a,int b){    return a<b?a:b;}inline void addedge(int u,int v){     edge[cnt].to=v;     edge[cnt].next=head[u];     head[u]=cnt++;}inline void makemap(int from,int to){     addedge(from,to);     addedge(to,from);}inline void DFS(int u,int father){     dp[u][1]=w[u];     for(int i=head[u];~i;i=edge[i].next)     {        int v=edge[i].to;        if(v==father)            continue;         else if(v!=father)         {             DFS(v,u);             for(int j=m;j>=2;j--)             {                 for(int k=1;k<j;k++)                 {                     dp[u][j]=max(dp[u][j],dp[v][k]+dp[u][j-k]);                 }             }         }     }     res=max(res,dp[u][m]);}int main(){     while(scanf("%d%d",&n,&m)!=EOF)     {         cnt=0;         memset(head,-1,sizeof(head));         memset(dp,0,sizeof(dp));         for(int i=0;i<n;i++)         {             scanf("%d",&w[i]);         }         for(int i=1;i<n;i++)         {            int u,v;             scanf("%d%d",&u,&v);             makemap(u,v);         }         res=0;         DFS(0,-1);         printf("%d\n",res);     }     return 0;}