poj2486 Apple Tree

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8672		Accepted: 2884

Description

Wshxzt is a lovely girl. She likes apple very much. One day HX takes her to an apple tree. There are N nodes in the tree. Each node has an amount of apples. Wshxzt starts her happy trip at one node. She can eat up all the apples in the nodes she reaches. HX is a kind guy. He knows that eating too many can make the lovely girl become fat. So he doesn’t allow Wshxzt to go more than K steps in the tree. It costs one step when she goes from one node to another adjacent node. Wshxzt likes apple very much. So she wants to eat as many as she can. Can you tell how many apples she can eat in at most K steps.

Input

There are several test cases in the input 
Each test case contains three parts. 
The first part is two numbers N K, whose meanings we have talked about just now. We denote the nodes by 1 2 ... N. Since it is a tree, each node can reach any other in only one route. (1<=N<=100, 0<=K<=200) 
The second part contains N integers (All integers are nonnegative and not bigger than 1000). The ith number is the amount of apples in Node i. 
The third part contains N-1 line. There are two numbers A,B in each line, meaning that Node A and Node B are adjacent. 
Input will be ended by the end of file. 

Note: Wshxzt starts at Node 1.

Output

For each test case, output the maximal numbers of apples Wshxzt can eat at a line.

Sample Input

2 1 
0 11
1 2
3 2
0 1 2
1 2
1 3

Sample Output

11

2

题意：给你一棵树，起点位置是1，树上的每个节点都有自己的价值，问你最多走k步能得到的节点的最大价值。

思路：首先容易想到状态方程dp[i][j]表示从i节点出发走j步所能得到的最大价值，但是这样定义状态会有一个问题，就是走j步后不一定会回到原来的点，这样就不能转移方程了，所以可以增加一维，用dp[i][j][0]表示i节点出发走j步并且最后回到i点的最大价值，用dp[i][j][1]表示i节点出发走j步并且最后不回到i点的最大价值，这样就容易转移了。

dp[i][j][0] = MAX (dp[i][j][0] , dp[i][j-k][0] + dp[son][k-2][0]);//从s出发，要回到s，需要多走两步s-t,t-s,分配给t子树k步，其他子树j-k步，都返回 dp[i][j]][1] = MAX(  dp[i][j][1] , dp[i][j-k][0] + dp[son][k-1][1]) ;//先遍历s的其他子树，回到s，遍历t子树，在当前子树t不返回，多走一步 dp[i][j][1] = MAX (dp[i][j][1] , dp[i][j-k][1] + dp[son][k-2][0]);//不回到s（去s的其他子树），在t子树返回，同样有多出两步

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
typedef long long ll;
#define maxn 105
int value[maxn],first[maxn],dp[maxn][2*maxn][2];
struct node{
    int to,next;
}e[2*maxn];
int n,k,vis[maxn];

void dfs(int u)
{
    int i,j,v,l;
    vis[u]=1;
    for(j=0;j<=k;j++){
        dp[u][j][0]=value[u];
        dp[u][j][1]=value[u]; //这里不管回不回到u点，初始值都是value[u],因为不移动也有value[u]
    }
    for(i=first[u];i!=-1;i=e[i].next){
        v=e[i].to;
        if(vis[v])continue;
        dfs(v);
        for(j=k;j>=1;j--){
            for(l=j;l>=1;l--){
                if(l>=2){
                    dp[u][j][0]=max(dp[u][j][0],dp[u][j-l][0]+dp[v][l-2][0]  );
                    dp[u][j][1]=max(dp[u][j][1],dp[u][j-l][1]+dp[v][l-2][0] );
                }
                dp[u][j][1]=max(dp[u][j][1],dp[u][j-l][0]+dp[v][l-1][1] );
            }
        }
    }
}


int main()
{
    int m,i,j,tot,u,v;
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        for(i=1;i<=n;i++){
            scanf("%d",&value[i]);
        }

        tot=0;
        memset(first,-1,sizeof(first));
        for(i=1;i<=n-1;i++){
            scanf("%d%d",&u,&v);
            tot++;
            e[tot].to=v;e[tot].next=first[u];
            first[u]=tot;

            tot++;
            e[tot].to=u;e[tot].next=first[v];
            first[v]=tot;

        }
        memset(dp,0,sizeof(dp));
        memset(vis,0,sizeof(vis));
        dfs(1);
        printf("%d
",max(dp[1][k][0],dp[1][k][1]) );
    }
    return 0;
}