POJ2385--Apple Catching(动态规划)
Each minute, one of the two apple trees drops an apple. Bessie, having much practice, can catch an apple if she is standing under a tree from which one falls. While Bessie can walk between the two trees quickly (in much less than a minute), she can stand under only one tree at any time. Moreover, cows do not get a lot of exercise, so she is not willing to walk back and forth between the trees endlessly (and thus misses some apples).
Apples fall (one each minute) for T (1 <= T <= 1,000) minutes. Bessie is willing to walk back and forth at most W (1 <= W <= 30) times. Given which tree will drop an apple each minute, determine the maximum number of apples which Bessie can catch. Bessie starts at tree 1.
Input
* Lines 2..T+1: 1 or 2: the tree that will drop an apple each minute.
Output
Sample Input
7 2
2
1
1
2
2
1
1
Sample Output
6
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;
int dp[][];
int apple[];
int main(){
int n,m;
cin>>n>>m;
for(int i=;i<=n;i++)
cin>>apple[i];
for(int i=;i<=n;i++){
for(int j=;j<=m;j++){
if(j==){
dp[i][j]=dp[i-][j];
}
else{
dp[i][j]=max(dp[i-][j],dp[i-][j-]);
}
if(j%+==apple[i])
dp[i][j]++;
}
}
int ans=dp[n][];
for(int i=;i<=m;i++){
if(ans<dp[n][i])
ans=dp[n][i];
}
cout<<ans<<endl;
return ;
}
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