POJ2385 Apple catching "DP"

Apple Catching

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8018		Accepted: 3922

Description

It is a little known fact, cows love apples. Farmer John has the apple trees (which is conveniently numbered 1 and 2) in the His field, each of the apples. Bessie cannot reach the apples when they is 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 hits the ground (and no one wants to eat bruised Apples). Bessie is a-a quick eater, so an apple she does catch was eaten in just a few seconds.

Each minute, one of the trees drops Apple. Bessie, has much practice, can catch an apple if she's standing under a tree from which one falls. While Bessie can walk between the both trees quickly (in much less than a minute), she can stand under only one tree at any Time. Moreover, cows do don't get a lot of exercise, so she's not willing to walk back and forth between the trees endlessly (and Thus misses some apples).

Apples Fall (one minute) for T (1 <= T <=) minutes. Bessie is willing to walk back and forth at most W (1 <= w <=) times. Given which tree would 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 would 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 both in a row from tree 1, then both in a row from tree 2, then both 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 has dropped, then moving to Tree 2 for the next T Wo, then returning back to Tree 1 for the final.

Source

Usaco 2004 NovemberTest Instructions: There is a cow, two trees, start the cow under the 1th tree, each second tree in two trees will fall an apple, the cow can move the most between the two Trees w times, for n seconds the maximum number of cows can receive apples. The problem: Recently done a lot of questions are about this cow, but also to wipe the floor, but also to walk the maze, is the sun ... This is going to pick up the apples again, it's a bunker. Cough, to get to the point, the state of the problem is dp[i][j] represents the first I-second move up to J step to the most apples. The state transition equation isDp[i][j] = max (Dp[i-1][j], dp[i][j-1]);

#include <stdio.h> #include <string.h>int dp[1010][35];bool arr[1010];int max (int a, int b) {    return a ; B? A:B;} int main () {    int T, W, I, J, tmp, ans = 0;    scanf ("%d%d", &t, &w);    for (i = 1; I <= T; ++i) {        scanf ("%d", &tmp);        Arr[i] = tmp-1;    }    if (arr[1]) dp[1][1] =1;    else dp[1][0] = 1;    ans = 1;    for (i = 2; I <= T; ++i) {        dp[i][0] = dp[i-1][0] +!arr[i];        for (j = 1; j <= W; ++j) {            dp[i][j] = max (Dp[i-1][j], dp[i][j-1]);            DP[I][J] + =! ((J & 1) ^ arr[i]);            if (Dp[i][j] > ans) ans = dp[i][j];        }    }    printf ("%d\n", ans);    return 0;}

POJ2385 Apple catching "DP"