zoj 3201 Tree of Tree 树形dp基础题——树形背包

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

3 110 20 300 10 23 210 20 300 10 2

Sample Output

3040

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Author: LIU, Yaoting

Source: ZOJ Monthly, May 2009

这份题解现在看起来感觉好幼稚，就是一个简单的树形背包啊(2016/10/25)：

题意是给出一棵树，每个点都有权值，现在需要求有k个结点的子树的最大权值。

dp[i][j]表示序号为i的根节点，形成的节点数为j的"子树"的权值和

首先题目中的子树跟我们平时的子树不一样，他的子树就是一块连通的点集,其中子树有根节点，假如选了某个节点，可以不选它的子节点。

状态转移方程：dp[x][all]=max{ dp[x][j]+dp[y][all-j] } 其中x表示根节点为x,       y表示新考虑的子节点

dp[i][j]表示序号为i的根节点，形成的节点数为j的"子树"的权值和

#include<cstdio>#include<iostream>#include<cmath>#include<cstring>#include<climits>#include<string>#include<algorithm>#include<queue>#include<set>#include<vector>#include<map>#include<stack>//typedef __int64 ll;//  %I64d//by yskyskyer123using namespace std;const int maxn=100+10;int a[maxn];int n,k;vector<int >nex[maxn];int dp[maxn][maxn];//dp[i][j]表示序号为i的根节点，形成的节点数为j的"子树"的权值和bool vis[maxn];void DP(int x){    dp[x][1]=a[x]; //首先，根节点为x，节点数为1的子树，只有1个节点，就是自身，权值和就是他自己的权值    for(int i=0;i<nex[x].size();i++)//一开始只考虑有节点x，慢慢一个一个加入x的子节点，进行动态规划。     {        int y=nex[x][i];        if(vis[y])  continue;        vis[y]=1;        DP(y);                                      for(int all=k;all>=1;all--)/*<span style="color:#ff0000;">状态转移方程：dp[x][all]=max{  dp[x][j]+dp[y][all-j]   } 其中x表示根节点为x,y表示新考虑的子节点</span>            （且已经考虑了一些子节点，每次i+1，dp[x][k]的值都会更新为考虑了新节点后的值）*/        /*考虑到计算dp[x][all]由dp[x][j]+dp[y][all-j]决定，而为了让dp[x][j]表示不考虑子节点nex[x][i]的情形。       all的顺序必须从k到1*/        {            for(int ori=1;ori<=all;ori++)//ori表示不考虑nex[x][i]时子树节点个数            {                dp[x][all]=max(dp[x][ori]+dp[y][all-ori ]   ,dp[x ][all]);            }        }    }}int main(){    int i,j,x,y;    while(~scanf("%d%d",&n,&k))    {        for(int i=0;i<n;i++)        {            nex[i].clear();            scanf("%d",&a[i]);            for(int j=1;j<=k;j++)            {                dp[i][j]=0;            }        }        for(int i=1;i<=n-1;i++)        {            scanf("%d%d",&x,&y);            nex[x].push_back(y);            nex[y].push_back(x);        }        memset(vis,0,sizeof vis);        vis[0]=1;        DP(0);        int ans=0;        for(int i=0;i<n;i++)            ans=max(ans,dp[i][k]);         printf("%d\n",ans);    }}