poj2385 Apple catching (DP state transfer equation derivation)

https://vjudge.net/problem/POJ-2385

Swipe the first question on the first day of the simple DP.

State: Dp[i][j] represents the maximum number of apples that are obtained by moving J times in the second. The key is to think about moving J times, judging by J's parity where the person is.

Thought for a long time, and finally a reference to their own ideas and the latest code 52050207

I am more than what he lacks is the state of dp[i][0], and everything behind it is based on this.

1#include <iostream>2#include <cstdio>3#include <queue>4#include <cstring>5#include <algorithm>6#include <cmath>7#include <Set>8 #defineINF 0x3f3f3f3f9typedefLong Longll;Ten using namespacestd; One inta[1010], dp[1010][1010]; A intMain () - { -Memset (DP,0,sizeof(DP)); the     intN, M; -CIN >> N >>m; -      for(inti =1; I <= N; i++){ -CIN >>A[i]; +     } -      for(inti =1; I <= N; i++){ +dp[i][0] = dp[i-1][0]; A         if(A[i] = =1){//apples on the left atdp[i][0]++; -              for(intj =1; J <= M; J + +){ -                 if(j&1)//man on the right. -DP[I][J] = max (dp[i-1][J], dp[i-1][j-1]); -                 Else//man on the left -DP[I][J] = max (dp[i-1][J], dp[i-1][j-1])+1;  in             } -         } to         Else{//the apples are on the right +              for(intj =1; J <= M; J + +){ -                 if(j&1)//man on the right. theDP[I][J] = max (dp[i-1][J], dp[i-1][j-1])+1;  *                 Else//man on the left $DP[I][J] = max (dp[i-1][J], dp[i-1][j-1]); Panax Notoginseng             } -         } the     } +     intMAXM =-INF; A      for(inti =0; I <= m; i++){ theMAXM =Max (MAXM, Dp[n][i]); +     }  -cout << MAXM <<Endl; $     return 0; $}

poj2385 Apple catching (DP state transfer equation derivation)