POJ 2385 Apple Catching

Apple Catching

Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 13283 Accepted: 6462

Description

It is a little known fact that cows love apples. Farmer John has two apple trees (which are conveniently numbered 1 and 2) in his field, each full of apples. Bessie cannot reach the apples when they are on the tree, so she must wait for them to fall. However, she must catch them in the air since the apples bruise when they hit the ground (and no one wants to eat bruised apples). Bessie is a quick eater, so an apple she does catch is eaten in just a few seconds.

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

• Line 1: Two space separated integers: T and W

• Lines 2..T+1: 1 or 2: the tree that will drop an apple each minute.

Output

• Line 1: The maximum number of apples Bessie can catch without walking more than W times.

Sample Input

7 2
2
1
1
2
2
1
1

Sample Output

6

#include<cstdio>#include<algorithm>using namespace std;int N, W;const int maxn = 1000+5;const int maxm = 30+5;int tree[maxn];int dp[maxn][maxn];void solve(){        if(tree[1] == 1){ //第一次在树一下站着, apple从第一棵树上落下             dp[1][0] = 1; //不用移动，apple+1             dp[1][1] = 0; //移动，apple = 0         }        else if(tree[1] == 2){ //苹果从第二棵树上落下             dp[1][0] = 0; //不移动 apple = 0             dp[1][1] = 1; //移动，apple+1         }        for(int i = 2; i <= N; i++)            for(int j = 0; j <= W; j++){                if(j == 0 ){ //当时间为i时，移动次数为0, 此时还站在树 1 的下面                     dp[i][j] = dp[i-1][j] + tree[i]%2; //如果apple从树一上落下，那么tree[i]%2 = 1, 可以接到苹果。                    continue;                           //如果从树二上落下，tree[i]%2 = 0, 不可以接到苹果                 }                 dp[i][j] = max(dp[i-1][j], dp[i-1][j-1]);//时间为i的时候，移动j次时获得的最多apple，                 if(j%2+1 == tree[i])  //如果本次(由上得出的只是移动不移动的情况，并没有判断时间为i时，apple从哪个树上落下)                                                //是站在第 i 棵树下，就会多收获一颗apple                     dp[i][j]++;            }        int res = 0;            for(int i = 0; i <= W; i++)            res = max(res, dp[N][i]);        printf("%d\n", res);}int main(){    scanf("%d%d", &N, &W);    for(int i = 1; i <= N; i++)        scanf("%d", &tree[i]);    solve();    return 0;}