poj 2385 Apple Catching 【暴力dp】

Apple Catching

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 9790 Accepted: 4763

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.

题意：给定两个树，初始你在第一颗树上下方。现在会有T个苹果在Ts时间里落下来（每秒落下一个苹果），已经给出第is落下的苹果属于哪一棵树。现在要求在两个树之间最多移动W次，问你能够最多可以接到苹果的数目。 

思路：设置dp[i][j][k]——第i秒在第k棵树下方且已经移动j次所能获取的最大苹果数目。

状态转移dp[i][j][k] = max(dp[i-1][j][k], dp[i-1][j-1][k%2+1])，当第i秒落下的苹果属于第k棵树时要加一。

注意j == 0时的处理。暴力转移就可以了。

AC代码：

#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <algorithm>#include <queue>#include <stack>#include <map>#include <set>#include <vector>#define INF 0x3f3f3f#define eps 1e-8#define MAXN (1000+1)#define MAXM (100000)#define Ri(a) scanf("%d", &a)#define Rl(a) scanf("%lld", &a)#define Rf(a) scanf("%lf", &a)#define Rs(a) scanf("%s", a)#define Pi(a) printf("%d\n", (a))#define Pf(a) printf("%.2lf\n", (a))#define Pl(a) printf("%lld\n", (a))#define Ps(a) printf("%s\n", (a))#define W(a) while(a--)#define CLR(a, b) memset(a, (b), sizeof(a))#define MOD 1000000007#define LL long long#define lson o<<1, l, mid#define rson o<<1|1, mid+1, r#define ll o<<1#define rr o<<1|1using namespace std;int num[MAXN];int dp[MAXN][40][3];int main(){    int T, W;    while(scanf("%d%d", &T, &W) != EOF)    {        CLR(dp, 0);        for(int i = 1; i <= T; i++)            Ri(num[i]);        for(int i = 1; i <= T; i++)        {            for(int j = 0; j <= W; j++)            {                for(int k = 1; k <= 2; k++)                {                    if(j == 0)                        dp[i][j][k] = dp[i-1][j][k];                    else                        dp[i][j][k] = max(dp[i-1][j][k], dp[i-1][j-1][(k%2+1)]);                    if(num[i] == k)                        dp[i][j][k]++;                }            }        }        Pi(max(dp[T][W][1], dp[T][W][2]));    }    return 0;}