zoukankan      html  css  js  c++  java
  • poj2385(dp)

    题目链接:http://poj.org/problem?id=2385

    Apple Catching
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 9372   Accepted: 4565

    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

    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

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <stack>
    #include <queue>
    #include <map>
    #include <set>
    #include <vector>
    #include <cmath>
    #include <algorithm>
    using namespace std;
    #define ll long long
    const double eps = 1e-6;
    const double pi = acos(-1.0);
    const int INF = 0x3f3f3f3f;
    const int MOD = 1000000007;
    
    int t,w,x;
    int f[1005][35],a[1005];
    
    int main ()
    {
        while (scanf ("%d%d",&t,&w)==2)
        {
            memset(f, 0,sizeof(f));
            for (int i=1; i<=t; i++)
            {
                scanf ("%d",&a[i]);
                f[i][0] = f[i-1][0]+(a[i]==1?1:0);
            }
            for (int j=1; j<=w; j++)
            {
                for (int i=1; i<=t; i++)
                {
                    f[i][j] = max(f[i-1][j], f[i-1][j-1])+(a[i]==(j%2+1)?1:0);
                }
            }
            printf ("%d
    ",f[t][w]);
        }
        return 0;
    }
    


     

  • 相关阅读:
    【vim】分割窗口、标签页与Quickfix窗口
    新手学cocos2dx,centos7下的安装过程
    外部排序,杀鸡焉用牛刀?
    5亿整数的大文件,怎么排?
    【Hadoop】HDFS
    你好,树
    写给博客园博客团队,md的预览在哪里?
    老菜鸟致青春,程序员应该选择java 还是 c#-
    高性能server分析
    高性能服务端漫谈
  • 原文地址:https://www.cnblogs.com/lytwajue/p/7001284.html
Copyright © 2011-2022 走看看