Dp:apple catching (POJ 2385)

　　　　　　　　　　　　

How Cows eat apples

The main problem: a cow called Bessie, can eat apples, and then there are two trees, apples on the tree will be dropped one minute, the cow one minute can be in two trees to eat apples (and do not eat on the ground), and then turn back only limited times W, ask you This is called Bessie cattle can eat up to how many apples

First we should be very easy to think that this must be done with DP, and then is to consider how to store the old value of the problem, because the tree has two trees, and then each round-trip state corresponds to different results, so we should use a two-dimensional matrix to store (the number of apples is not important, because we only calculate the last, The front can be thrown away).

And how should the state equation be written? It is also very simple to define a state of dp[i][j] that represents the maximum number of times a tree can be eaten by the time it is left, and the cow can walk from a tree next to it.

So the state transfer equation is natural.

Dp[i][j]=max (dp[i][j],dp[i-1][j+1]) +1 (j<w && i+1>w-j)

dp[i][j]=dp[i][j]+1; (j=w)

　　Also note that there are some states are meaningless, typical is the time of i+1<w-j (you think that the number of minutes the Apple dropped more than the number of cows ran over, how possible!) ）

　　

1#include <stdio.h>2#include <stdlib.h>3 #defineMAX (a) (a) > (b)? (a):(B)4 5 Static inttree[2][ to];6 7 intMainvoid)8 {9     intT, W, I, K, which_tree, ans =0;Ten      while(~SCANF ("%d%d", &t, &W)) One     { A          for(i =0; i < T; i++) -         { -scanf"%d", &which_tree); theTree[which_tree-1][w]++; -              for(k = W-1; K >=0&& i +1> w-k; k--) -Tree[which_tree-1][K] = -MAX (tree[! ( Which_tree-1)][k +1] +1, Tree[which_tree-1][K] +1); +         } -          for(i =0; I <2; i++) +              for(k =0; K <= W; k++) AAns =MAX (Tree[i][k], ans); atprintf"%d", ans); -     } -  -     return 0; -}

Dp:apple catching (POJ 2385)