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. linux系统基础(二)

    磁盘管理(一) Linux设备认识 /dev/cdrom /dev/sr0 /dev/mouse /dev/sda /dev/hda IDE硬盘(支持4块):hd(a-d) [非IDE硬盘]SCSI硬 ...

  2. error: variable '__this_module' has initializer but incomplete type错误解决

    版权所有,转载必须说明转自 http://my.csdn.net/weiqing1981127 原创作者:南京邮电大学  通信与信息系统专业 研二 魏清 问题描述:使用SAM9X25  内核版本是2. ...

  3. js url传值中文乱码之解决之道

    在websphere 中使用的是url=encodeURI(encodeURI(url)); //用了2次encodeURI 测试成功,第一次转换没有尝试, 处理方法一. js 程序代码:url=en ...

  4. C#/Access-数据库获取自动编号的最大值

    //conStrSQL你改成你的access,我这里用的SQL2005string conStrSQL = "Data Source=xx.xx.xx.xx;Initial Catalog= ...

  5. [Sciter系列] MFC下的Sciter–1.创建工程框架

      Sciter SDK中提供的Win32下例程很多,唯独使用很多(对我个人而言)的MFC框架下Sciter程序的构建讲的很少,虽然MFC有这样那样的诟病,但是不可否认的是编写一般的小项目,这仍然是大 ...

  6. 查看Linux主机CPU及内存信息

    查看CPU信息(型号)  # cat /proc/cpuinfo | grep name | cut -f2 -d: | uniq -c        8  Intel(R) Xeon(R) CPU ...

  7. [React Native + Firebase] React Native: Real time database with Firebase -- setup & CRUD

    Install: npm i --save firebase // v3.2.1 Config Firebase: First we need to require Firebase: import ...

  8. shell重定向调试信息

    shell重定向调试信息 fulinux ******************************************************************************* ...

  9. android123 zhihuibeijing 新闻中心-新闻 页签 ViewPagerIndicator实现

    ## ViewPagerIndicator ## 使用导入ViewPagerIndicator库的方式相当于可以改源码,打包编译Eclips可以自动完成. ViewPager指针项目,在使用ViewP ...

  10. 深入理解计算机系统第二版习题解答CSAPP 2.16

    填写下表,说明不同移位运算对单字节数的影响. x x<<3 x>>2(逻辑) x>>2(算术) 十六进制 二进制 二进制 十六进制 二进制 十六进制 二进制 十六进 ...