POJ 2486 Apple Tree

                                                                                                                                               Apple Tree

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7120		Accepted: 2370

Description

Wshxzt is a lovely girl. She likes Apple very much. One day HX takes she to an apple tree. There is N nodes in the tree. Each node has a amount of apples. Wshxzt starts her happy trips 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 many apples she can eat in to the most K steps.

Input

There is several test cases in the input
Each test case contains three parts.
The first part is a numbers N K, whose meanings we have a talked about just now. We denote the nodes by 1 2 ... N. Since It is a tree with each node can reach any and other on only one route. (1<=n<=100, 0<=k<=200)
The second part contains N integers (all integers is 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 is numbers a, A, a, and node B are adjacent, meaning.
Input would 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 0 111 23 20 1 21 21 3

Sample Output

112

Source

Although the topic is to look at the problem-solving report, but it is finally over. It's hard for me very very very long long time

Dp[a][b][0]: Node A go b step back to Node A

DP[A][B][1]: Node A step b does not go back to Node A

#include <iostream> #include <cstdio> #include <cmath> #include <cstring> #include < algorithm> #include <cstdlib> #include <queue> #include <set> #define N 210#define M 110using namespace std;struct num{int x,y,next;}    A[2*n];int b[n],top,n,m;int Dp[n][n][2],temp[n][n][2];int Main () {//freopen ("Data.txt", "R", stdin);    void addeage (int x,int y);    void Dfs (int u,int FA);        while (scanf ("%d%d", &n,&m)!=eof) {memset (dp,0,sizeof (DP));            for (int i=1;i<=n;i++) {int x;            scanf ("%d", &x);            Dp[i][0][0] = x;        DP[I][0][1] = x;        } memset (b,-1,sizeof (b));        Top = 0;            for (int i=1;i<=n-1;i++) {int x, y;            scanf ("%d%d", &x,&y);            Addeage (x, y);        Addeage (Y,X);        } dfs (1,-1);        int ans = 0;        for (int i=0;i<=m;i++) {ans = max (ans,dp[1][i][0]);        }printf ("%d\n", ans); } return 0;}    void addeage (int x,int y) {a[top].y = y;    A[top].next = B[x]; B[X] = top++;}        void Dfs (int u,int fa) {for (int i=b[u];i!=-1;i=a[i].next) {int v = a[i].y;        if (V==FA) {continue;        } dfs (V,U); for (int i=1;i<=n;i++) {for (int j=0;j<=m;j++) {temp[i][j][0] = Dp[i][j]                [0];            TEMP[I][J][1] = dp[i][j][1]; }} for (int i=1;i<=m;i++) {for (int j=1;j<=i;j++) {dp[u][                I][0] = max (dp[u][i][0],temp[u][i-j][1]+temp[v][j-1][0]);                    if (j>=2) {dp[u][i][0] = max (dp[u][i][0],temp[v][j-2][1]+temp[u][i-j][0]);                DP[U][I][1] = max (dp[u][i][1],temp[u][i-j][1]+temp[v][j-2][1]); }            }        }    }}