[Poj 2486] Apple Tree 树形DP
Apple Tree
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 9069 Accepted: 3016
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
Source
POJ Contest,Author:magicpig@ZSU
题目链接：http://poj.org/problem?id=2486；
题意：
一颗树，n个点（1-n），n-1条边，每个点上有一个权值，求从1出发，走V步，最多能遍历到的权值；
思路：
树形ＤＰ；
dp[i][j][0/1]表示在ｉ的子树下走ｊ步（回到ｉ点用１表示）；
dp[x][j][0]=max(dp[x][j][0],dp[x][j-t][1]+dp[y][t-1][0]);／／先回ｘ点（－１）；
dp[x][j][0]=max(dp[x][j][0],dp[x][j-t][0]+dp[y][t-2][1]);／／先回ｙ点－》ｘ点（－２）；
dp[x][j][1]=max(dp[x][j][1],dp[x][j-t][1]+dp[y][t-2][1]);//回到ｘ点必回ｙ点（－２）代码：
#includeint y=lin[x][i];
if(y==pre) continue;//不搜回边
dfs(y,x);
for(int j=k;j>=1;j--)
for(int t=1;t<=j;t++)
{
dp[x][j][0]=max(dp[x][j][0],dp[x][j-t][1]+dp[y][t-1][0]);//先回ｘ点（－１）；
dp[x][j][0]=max(dp[x][j][0],dp[x][j-t][0]+dp[y][t-2][1]);//先回ｙ点－》ｘ点（－２）；
dp[x][j][1]=max(dp[x][j][1],dp[x][j-t][1]+dp[y][t-2][1]);//回到ｘ点必回ｙ点（－２）
}
}
}
int main()
{
while(scanf("%d%d",&n,&k)!=EOF)
{
memset(dp,0,sizeof(dp));
int aa,bb;
for(int i=1;i<=n;i++)
{
lin[i].clear();
scanf("%d",&aa);
for(int j=0;j<=k;j++)
{
dp[i][j][0]=dp[i][j][1]=aa;／／初始化，至少为本节点的值
}
}
for(int i=1;iscanf("%d%d",&aa,&bb);
lin[aa].push_back(bb);
lin[bb].push_back(aa);
}
dfs(1,0);//树形dp，存的双向边，所以默认１
为根节点；
printf("%d\n",max(dp[1][k][0],dp[1][k][1]));//答案从１走，回到１与不会的答案；
}
}