Apple Catching POJ

Apple Catching

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 11916 Accepted: 5772

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 22112211

Sample Output

6

Hint

INPUT DETAILS: 

Seven apples fall - one from tree 2, then two in a row from tree 1, then two in a row from tree 2, then two in a row from tree 1. Bessie is willing to walk from one tree to the other twice. 

OUTPUT DETAILS: 

Bessie can catch six apples by staying under tree 1 until the first two have dropped, then moving to tree 2 for the next two, then returning back to tree 1 for the final two.

Source

USACO 2004 November

题目大意：总共T秒，2棵树，每秒在一棵树上会掉一个苹果，最多可以回到某棵树K次，问接到苹果最大的数目

思路：  dp[i][j][k] 第i秒，位于j树，跳跃k次时的最大苹果数。k>= 1时 dp[i + 1][j][k] = dp[i][j][k] + dp[i][j ^ 1][k - 1]；k==0时 dp[i + 1][j][k] = dp[i][j][k]。之后在判断下第i+1秒处是否可以接到苹果。

#include <iostream>#include <cstdio>using namespace std;int dp[1005][3][35];int a[1005];int main(){    ///dp[i][j][k] 第i秒，位于j树，跳跃k次时的最大苹果数    ///dp[i + 1][j][k] = dp[i][j][k] + dp[i][j ^ 1][k - 1];    int T, W;    scanf("%d%d", &T, &W);    for(int i = 1; i <= T; ++i) {        scanf("%d", &a[i]);    }    dp[1][a[1] - 1][0] = 1;    for(int i = 1; i < T; ++i) {        for(int j = 0; j <= 1; ++j) {            for(int k = 0; k <= W; ++k) {                if(k >= 1) {                    dp[i + 1][j][k] = max(dp[i][j][k], dp[i][j ^ 1][k - 1]);                } else {                    dp[i + 1][j][k] = dp[i][j][k];                }                if(j == (a[i + 1] - 1)) {                    dp[i + 1][j][k]++;                }                //cout << "i:" << i + 1 << "    j:" << j + 1 << "     k:" << k << "    ans:" << dp[i + 1][j][k] << endl;            }        }    }    int ans = 0;    for(int k = 0; k <= W; ++k) {        ans = max(ans, dp[T][1][k]);        ans = max(ans, dp[T][0][k]);    }    printf("%d\n", ans);    return 0;}