ZOJ 3201 Tree of Tree

树形背包。。。

Tree of Tree Time Limit: 1000MS   Memory Limit: 32768KB   64bit IO Format: %lld & %llu [Submit]   [Go Back]   [Status]   Description 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 Source ZOJ Monthly, May 2009 [Submit]   [Go Back]   [Status]

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int INF=0x3f3f3f3f;

int Adj[200],Size;

struct E
{
int to,next;
}Edge[40000];

void init()
{
Size=0;
memset(Adj,-1,sizeof(Adj));
}

void Add_Edge(int u,int v)
{
Edge[Size].to=v;
Edge[Size].next=Adj[u];
Adj[u]=Size++;
}

int n,K,dp[200][200];

void dfs(int f,int u)
{
for(int i=Adj[u];~i;i=Edge[i].next)
{
int v=Edge[i].to;
if(v==f) continue;
dfs(u,v);
for(int j=K;j>=1;j--)
{
for(int k=1;k<j;k++)
{
dp[u][j]=max(dp[u][j],dp[u][j-k]+dp[v][k]);
}
}
}
}

int main()
{
while(scanf("%d%d",&n,&K)!=EOF)
{
init();
memset(dp,0,sizeof(dp));
for(int i=0;i<n;i++) scanf("%d",&dp[i][1]);
for(int i=0;i<n-1;i++)
{
int u,v;
scanf("%d%d",&u,&v);
Add_Edge(u,v);
Add_Edge(v,u);
}
dfs(-1,0);
int ans=-INF;
for(int i=0;i<n;i++)
ans=max(ans,dp[i][K]);
printf("%d\n",ans);
}
return 0;
}