POJ 2385 Apple Catching (DP)

Description

It isA little known fact that cows love apples. Farmer John has apple trees (which is conveniently numbered1and2)inchHis field, each of the full of apples. Bessie cannot reach the apples when they is on the tree, so she must wait forthem to fall. However, she mustCatchtheminchThe air since the apples bruise when they hits the ground (and no one wants to eat bruised apples). Bessie isA quick eater, so an apple she doesCatch  isEateninchjust a few seconds. Each minute, one of the trees drops Apple. Bessie, having much practice, canCatchAn appleifShe isStanding under a tree fromWhich one falls. While Bessie can walk between the trees quickly (inchMuch less than a minute), she can stand under only one tree at any time. Moreover, cows DoNotGetA lot of exercise, so she isNot willing to walk back and forth between the trees endlessly (and thus misses some apples ). Apples Fall (one each minute) forT1<= T <=1, the) minutes. Bessie isWilling to walk back and forth at the most W (1<= W <= -) times. Given which tree would drop an apple each minute, determine the maximum number of apples which Bessie canCatch. Bessie starts at tree1.

Input

12.. t+112: The tree that would drop an apple each minute.

Output

1 catch without walking more than W times.

Sample Input

7 2 2 1 1 2 2 1 1

Sample Output

6

Hint

INPUT details:seven apples fall-One fromTree2, then bothinchA row fromTree1, then bothinchA row fromTree2, then bothinchA row fromTree1. Bessie isWilling to walk fromOne tree to the other twice. OUTPUT Details:bessie canCatchSix apples by staying under tree1Until the first and dropped, then moving to tree2  forThe next, then returning back to the tree1  forThe final.

Source

Usaco 2004 November   Set Dp[i][j] indicates the maximum value of the apple when walking J-step when I find the first Apple. to initialize first, see the Code Dp[i][j]=max (dp[i-1][j],dp[i-1][j-1]), indicating the maximum value of walking or not walking. Then judge whether you can dp[i][j]++. Finally find the maximum value

1#include <iostream>2#include <cstdio>3#include <cstring>4#include <algorithm>5#include <stdlib.h>6#include <cmath>7 using namespacestd;8 #defineW 369 #defineN 1006Ten intDp[n][w]; One intn,w; A intA[n]; - intMain () - { the      while(SCANF ("%d%d", &n,&w) = =2){ -         //int sum=0; -          for(intI=1; i<=n;i++){ -scanf"%d",&a[i]); +         } -Memset (DP,0,sizeof(DP)); +         if(a[1]==1){ Adp[1][0]=1; atdp[1][1]=0; -         } -         if(a[1]==2){ -dp[1][0]=0; -dp[1][1]=1; -         } in  -          for(intI=2; i<=n;i++){ to              for(intj=0; j<=w;j++){ +                 if(j==0){ -dp[i][j]=dp[i-1][j]+ (j%2+1==a[i]); the                     Continue; *                 } $Dp[i][j]=max (dp[i-1][j],dp[i-1][j-1]);Panax Notoginseng                 if(j%2+1==A[i]) { -dp[i][j]++; the                 } +             } A         } the         intans=dp[n][0]; +          for(intI=1; i<=w;i++){ -ans=Max (ans,dp[n][i]); $         } $printf"%d\n", ans); -  -     } the     return 0; -}

View Code

There is another way:

Set Dp[i][j] means that when I find the apple, I walk up to the max of J-Step apples

You can run the

Before i-1 minutes up to walk J times

Up to J-1 times before I-1 min.

These two states are moving over.

Note that the second type of transfer J can choose to walk or not to go. Because it's going to be at most J times.

With a tree-shaped DP that I've done before.

1#include <iostream>2#include <cstdio>3#include <cstring>4#include <algorithm>5#include <stdlib.h>6#include <cmath>7 using namespacestd;8 #defineW 369 #defineN 1006Ten intDp[n][w]; One intn,w; A intA[n]; - intMain () - { the      while(SCANF ("%d%d", &n,&w) = =2){ -         //int sum=0; -          for(intI=1; i<=n;i++){ -scanf"%d",&a[i]); +         } -Memset (DP,0,sizeof(DP)); +         if(a[1]==1) dp[1][0]=1; Adp[1][1]=1; at          for(intI=2; i<=n;i++){ -              for(intj=0; j<=w;j++){ -                 if(j==0){ -dp[i][j]=dp[i-1][j]+ (j%2+1==a[i]); -                     Continue; -                 } in  -Dp[i][j]=max (dp[i][j],dp[i-1][j]+ (j%2+1==a[i])); toDp[i][j]=max (dp[i][j],dp[i-1][j-1]+ (j%2==a[i])); +Dp[i][j]=max (dp[i][j],dp[i-1][j-1]+ (j%2+1==a[i])); -             } the         } *printf"%d\n", Dp[n][w]); $     }Panax Notoginseng     return 0; -}

View Code

POJ 2385 Apple Catching (DP)