POJ 2385 Apple catching DP

http://poj.org/problem?id=2385

Serie A: There are two apples, and there will be an apple falling from a tree on the clock. You have to stand under this tree to be able to get the apples falling from this tree, you can walk back and forth between the two trees, and do not spend time, but do not walk more than w times, now tell you every minute from which tree will fall from the Apple, ask you how many can be followed, your starting position is under the first tree.

Idea: Use Dp[i][j][k] to denote the first minute. had gone back and forth for J times, and now under the K-Tree (k=0) under the first tree. K=1 said that under the second tree), the maximum number of apples that can be received, then suppose that I minute, is the first lesson the tree falls off the apple. So easy to think of Dp[i]][j][0]=max (dp[i-1][j][0]+1,dp[i-1][j-1][1]+1), because probably before the first tree, then do not need to move around, you can pick another apple, or before the second tree, Then it is necessary to move the talent to the first tree, and then another Apple; Dp[i][j][1]=dp[i-1][j][1], thought according to Greedy thought, it is impossible to know the first tree to fall apple. He also walked under the first tree under the second tree, so K=1 's situation was only introduced by Dp[i-1][j][1. So in the same vein, we can get the first minute, the second tree falls off the apple, and here is no longer described. If the hypothesis is not clear, look at the following code.

It is important to note that the final answer is not necessarily the maximum number of moves, such as all the first apple tree in the apple, so you can get the most apples without moving, so you have to find the maximum value in all cases.

#include <cstdio> #include <cstring> #include <iostream>using namespace Std;int dp[1010][35][2];int A    [1010];int Main () {int n,m;        while (cin>>n>>m) {memset (dp,0,sizeof (DP));        for (int i=1;i<=n;i++) cin>>a[i];                    for (int i=1;i<=n;i++) {for (int j=0;j<=m;j++) if (j<=i) {if (a[i]==1) {                    if (j) Dp[i][j][0]=max (dp[i-1][j][0]+1,dp[i-1][j-1][1]+1);                    else dp[i][j][0]=dp[i-1][j][0]+1;                DP[I][J][1]=DP[I-1][J][1];                    } else{dp[i][j][0]=dp[i-1][j][0];                    if (j) Dp[i][j][1]=max (dp[i-1][j][1]+1,dp[i-1][j-1][0]+1);                else dp[i][j][1]=dp[i-1][j][1]+1;        }}} int ans=0;            for (int i=0;i<=m;i++) {Ans=max (ans,dp[n][i][0]);        Ans=max (ans,dp[n][i][1]); } cout<<ans<<endl; } return 0;}

POJ 2385 Apple catching DP