链接地址:http://bailian.openjudge.cn/practice/1928

题目:

总时间限制:
1000ms
内存限制:
65536kB
描述
Mr. Robinson and his pet monkey Dodo love peanuts very much. One day while they were having a walk on a country road, Dodo found a sign by the road, pasted with a small piece of paper, saying "Free Peanuts Here! " You can imagine how happy Mr. Robinson and Dodo were.

There
was a peanut field on one side of the road. The peanuts were planted
on the intersecting points of a grid as shown in Figure-1. At each
point, there are either zero or more peanuts. For example, in Figure-2,
only four points have more than zero peanuts, and the numbers are 15,
13, 9 and 7 respectively. One could only walk from an intersection
point to one of the four adjacent points, taking one unit of time. It
also takes one unit of time to do one of the following: to walk from
the road to the field, to walk from the field to the road, or pick
peanuts on a point.

According
to Mr. Robinson's requirement, Dodo should go to the plant with the
most peanuts first. After picking them, he should then go to the next
plant with the most peanuts, and so on. Mr. Robinson was not so patient
as to wait for Dodo to pick all the peanuts and he asked Dodo to return
to the road in a certain period of time. For example, Dodo could pick
37 peanuts within 21 units of time in the situation given in Figure-2.

Your
task is, given the distribution of the peanuts and a certain period of
time, tell how many peanuts Dodo could pick. You can assume that each
point contains a different amount of peanuts, except 0, which may appear
more than once.

输入
The first line of input contains the test case number T (1 <= T
<= 20). For each test case, the first line contains three integers,
M, N and K (1 <= M, N <= 50, 0 <= K <= 20000). Each of the
following M lines contain N integers. None of the integers will exceed
3000. (M * N) describes the peanut field. The j-th integer X in the
i-th line means there are X peanuts on the point (i, j). K means Dodo
must return to the road in K units of time.
输出
For each test case, print one line containing the amount of peanuts Dodo can pick.
样例输入
2
6 7 21
0 0 0 0 0 0 0
0 0 0 0 13 0 0
0 0 0 0 0 0 7
0 15 0 0 0 0 0
0 0 0 9 0 0 0
0 0 0 0 0 0 0
6 7 20
0 0 0 0 0 0 0
0 0 0 0 13 0 0
0 0 0 0 0 0 7
0 15 0 0 0 0 0
0 0 0 9 0 0 0
0 0 0 0 0 0 0
样例输出
37
28
来源
Beijing 2004 Preliminary@POJ

思路:

每次采摘前计算是否能够采摘,模拟题

代码:

 #include <iostream>
#include <cstdlib>
using namespace std; int main()
{
int t;
cin>>t;
while(t--)
{
int m,n,k;
cin>>m>>n>>k;
int *arr = new int[m*n];
for(int i = ; i < m; i++)
{
for(int j = ; j < n; j++)
{
cin>>arr[i * n + j];
}
}
int sum = ,ni=,nj;
int max,maxi,maxj;
k -= ;
int flag = ;
do
{
max = -;
for(int i = ; i < m; i++)
{
for(int j = ; j < n; j++)
{
if(max < arr[i * n + j])
{
max = arr[i * n + j];
maxi = i;
maxj = j;
}
}
}
if(flag) {nj = maxj;flag = ;}
int step = abs(maxi - ni) + abs(maxj - nj) + ;
if(step + maxi > k) break;
else
{
k -= step;
sum += max;
ni = maxi;
nj = maxj;
arr[maxi * n + maxj] = ;
} }while();
cout<<sum<<endl;
delete [] arr;
}
return ;
}

OpenJudge / Poj 1928 The Peanuts C++的更多相关文章

  1. OpenJudge / Poj 2141 Message Decowding

    1.链接地址: http://poj.org/problem?id=2141 http://bailian.openjudge.cn/practice/2141/ 2.题目: Message Deco ...

  2. OpenJudge/Poj 2105 IP Address

    1.链接地址: http://poj.org/problem?id=2105 http://bailian.openjudge.cn/practice/2105 2.题目: IP Address Ti ...

  3. OpenJudge/Poj 2027 No Brainer

    1.链接地址: http://bailian.openjudge.cn/practice/2027 http://poj.org/problem?id=2027 2.题目: 总Time Limit: ...

  4. OpenJudge/Poj 2013 Symmetric Order

    1.链接地址: http://bailian.openjudge.cn/practice/2013 http://poj.org/problem?id=2013 2.题目: Symmetric Ord ...

  5. OpenJudge/Poj 1088 滑雪

    1.链接地址: bailian.openjudge.cn/practice/1088 http://poj.org/problem?id=1088 2.题目: 总Time Limit: 1000ms ...

  6. OpenJudge/Poj 2001 Shortest Prefixes

    1.链接地址: http://bailian.openjudge.cn/practice/2001 http://poj.org/problem?id=2001 2.题目: Shortest Pref ...

  7. OpenJudge/Poj 2000 Gold Coins

    1.链接地址: http://bailian.openjudge.cn/practice/2000 http://poj.org/problem?id=2000 2.题目: 总Time Limit: ...

  8. OpenJudge/Poj 1936 All in All

    1.链接地址: http://poj.org/problem?id=1936 http://bailian.openjudge.cn/practice/1936 2.题目: All in All Ti ...

  9. OpenJudge/Poj 1661 帮助 Jimmy

    1.链接地址: bailian.openjudge.cn/practice/1661 http://poj.org/problem?id=1661 2.题目: 总Time Limit: 1000ms ...

随机推荐

  1. 解决下载Android Build-tools 19.1.0失败

    准备从Eclipse转到Android Studio了.今天尝试Android Studio的时候,被它提醒我SDK的Android Build-tools版本过低,需要升级. 于是打开Android ...

  2. Tomcat部署web应用的三种方式

    原文:http://my.oschina.net/sunchp/blog/90235 一:相关概念 CATALINA_HOME:tomcat安装目录 CATALINA_BASE:tomcat工作目录 ...

  3. 如何编写程序设置Android来电铃声

    我们在拿到新手机后通常会为其设置来年铃声,那么怎样通过代码来设置Android来电铃声,本文就为大家实例讲解下. 1.如果读到的是音频文件路径,需要先将音乐文件插入到多媒体库. Java代码 //设置 ...

  4. android115 自定义控件

    布局: <RelativeLayout xmlns:android="http://schemas.android.com/apk/res/android" xmlns:to ...

  5. C_数据结构_链表的链式实现

    传统的链表不能实现数据和链表的分离,一旦数据改变则链表就不能用了,就要重新开发. 如上说示:外层是Teacher,里面小的是node. #ifndef _MYLINKLIST_H_ #define _ ...

  6. UNIX标准化及实现之标准之间的冲突

    就整体而言,这些不同的标准之间配合得相当好.但是我们也很关注它们之间的差别,特别是ISO C标准和POSIX.1之间的差别. ISO C定义了函数clock,它返回进程使用的CPU时间,返回值类型是c ...

  7. endif、endforeach

    <?php if ($a == 5): ?>  <div>等于5</div><?php elseif ($a == 6): ?>  <div> ...

  8. C# 之 Excel 导入一列中既有汉字又有数字:数字可以正常导入,汉字导入为空

    今天在做一个Excel导入功能,一切开发就绪,数据可以成功导入.导入后检查数据库发现有一列既有汉字又有数字,数字正常导入,汉字为空.但是前面同样既有汉字又有数字的列可以导入成功. 查看excel 源文 ...

  9. android开发之路01

    一.android系统被分为4个层次:1.最下层的是linux核心,包括多个驱动程序,提供了操作系统应该具备的核心功能:2.在linux核心之上,包括两个部分,一部分是Android Runtime( ...

  10. 修改UILabel的行间距

    在iOS开发中  有时候为了调整一些UI效果  我们需要调整UILabel之间的行间距: contentLabel.text:label上显示的文字内容; 5:label行间距; contentLab ...