Problem Statement

We have a grid with H rows and W columns. The square at the i-th row and the j-th column will be called Square (i,j).

The integers from 1 through H×W are written throughout the grid, and the integer written in Square (i,j) is Ai,j.

You, a magical girl, can teleport a piece placed on Square (i,j) to Square (x,y) by consuming |xi|+|yj| magic points.

You now have to take Q practical tests of your ability as a magical girl.

The i-th test will be conducted as follows:

  • Initially, a piece is placed on the square where the integer Li is written.

  • Let x be the integer written in the square occupied by the piece. Repeatedly move the piece to the square where the integer x+D is written, as long as x is not Ri. The test ends when x=Ri.

  • Here, it is guaranteed that RiLi is a multiple of D.

For each test, find the sum of magic points consumed during that test.

Constraints

  • 1≤H,W≤300
  • 1≤DH×W
  • 1≤Ai,jH×W
  • Ai,jAx,y((i,j)≠(x,y))
  • 1≤Q≤105
  • 1≤LiRiH×W
  • (RiLi) is a multiple of D.

Input

Input is given from Standard Input in the following format:

H W D
A1,1 A1,2 A1,W
:
AH,1 AH,2 AH,W
Q
L1 R1
:
LQ RQ

Output

For each test, print the sum of magic points consumed during that test.

Output should be in the order the tests are conducted.

Sample Input 1

3 3 2
1 4 3
2 5 7
8 9 6
1
4 8

Sample Output 1

5
  • 4 is written in Square (1,2).

  • 6 is written in Square (3,3).

  • 8 is written in Square (3,1).

Thus, the sum of magic points consumed during the first test is (|3−1|+|3−2|)+(|3−3|+|1−3|)=5.

Sample Input 2

4 2 3
3 7
1 4
5 2
6 8
2
2 2
2 2

Sample Output 2

0
0

Note that there may be a test where the piece is not moved at all, and there may be multiple identical tests.

Sample Input 3

5 5 4
13 25 7 15 17
16 22 20 2 9
14 11 12 1 19
10 6 23 8 18
3 21 5 24 4
3
13 13
2 10
13 13

Sample Output 3

0
5
0

Q最高100000,如果每次都要计算,会超时,先记录前缀和,直接用前缀和相减就好了。
代码:
#include <bits/stdc++.h>///int * int = int
using namespace std;
int h,w,d,s[][],q,a,b;
int x[],y[];
int dp[];
int main()
{
scanf("%d%d%d",&h,&w,&d);
for(int i = ;i < h;i ++)
{
for(int j = ;j < w;j ++)
{
scanf("%d",&s[i][j]);
x[s[i][j]] = i;
y[s[i][j]] = j;
}
}
for(int i = + d;i <= h * w;i ++)
{
dp[i] = dp[i - d] + abs(x[i] - x[i - d]) + abs(y[i] - y[i - d]);
}
scanf("%d",&q);
for(int i = ;i < q;i ++)
{
scanf("%d%d",&a,&b);
printf("%lld\n",abs(dp[a] - dp[b]));
}
}

AtCoder Beginner Contest 089 D - Practical Skill Test的更多相关文章

  1. AtCoder Beginner Contest 089完整题解

    A - Grouping 2 Time limit : 2sec / Memory limit : 256MB Score : 100 points Problem Statement There a ...

  2. AtCoder Beginner Contest 100 2018/06/16

    A - Happy Birthday! Time limit : 2sec / Memory limit : 1000MB Score: 100 points Problem Statement E8 ...

  3. AtCoder Beginner Contest 052

    没看到Beginner,然后就做啊做,发现A,B太简单了...然后想想做完算了..没想到C卡了一下,然后还是做出来了.D的话瞎想了一下,然后感觉也没问题.假装all kill.2333 AtCoder ...

  4. AtCoder Beginner Contest 053 ABCD题

    A - ABC/ARC Time limit : 2sec / Memory limit : 256MB Score : 100 points Problem Statement Smeke has ...

  5. AtCoder Beginner Contest 136

    AtCoder Beginner Contest 136 题目链接 A - +-x 直接取\(max\)即可. Code #include <bits/stdc++.h> using na ...

  6. AtCoder Beginner Contest 137 F

    AtCoder Beginner Contest 137 F 数论鬼题(虽然不算特别数论) 希望你在浏览这篇题解前已经知道了费马小定理 利用用费马小定理构造函数\(g(x)=(x-i)^{P-1}\) ...

  7. AtCoder Beginner Contest 076

    A - Rating Goal Time limit : 2sec / Memory limit : 256MB Score : 100 points Problem Statement Takaha ...

  8. AtCoder Beginner Contest 079 D - Wall【Warshall Floyd algorithm】

    AtCoder Beginner Contest 079 D - Wall Warshall Floyd 最短路....先枚举 k #include<iostream> #include& ...

  9. AtCoder Beginner Contest 064 D - Insertion

    AtCoder Beginner Contest 064 D - Insertion Problem Statement You are given a string S of length N co ...

随机推荐

  1. php 防盗链

    防盗链的技术已经很普遍了,有些网站不喜欢自己的图片被别的网站直接复制使用,便使用了防盗链的技术,这样别人在直接复制使用网站图片时,图片便会按照程序的设定不显示或显示防盗链等字样. 使用了防盗链技术,不 ...

  2. 一、基础篇--1.1Java基础-Exception、Error、RuntimeException与一般异常有何异同

    Throwable.Error.Exception.RuntimeException 关系如下类图所示: Throwable: Throwable类是java语言中所有错误或者异常的超类.它的两个子类 ...

  3. leetcode-mid-array-sorting and searching - 215 Kth Largest Element in an Array

    mycode  77.39% class Solution(object): def findKthLargest(self, nums, k): """ :type n ...

  4. log() exp()函数

    1 对数函数表示法 import numpy as np import math print('输出自然底数e:',math.e) # np表示法 # np.log()是以e为底的自然对数 print ...

  5. lambda一些查询语句

    <!--得分数据结构-->1 <Score> <studentid>1</studentid> <courseid>1</course ...

  6. C 语言的函数 - 内存分析

    函数基本概念 Linux 中,函数在内存的代码段(code 区),地址比较靠前. 函数定义 C 语言中,函数有三个要素:入参.返回值.函数名,缺一不可.函数使用前必须先声明,或者在使用之前定义. 函数 ...

  7. 牛客练习赛46 E 华华和奕奕学物理 (树状数组)

    https://ac.nowcoder.com/acm/contest/894/E 一开始写了一个简单的模拟 通过率只有5%...... 看题解真的理解了好久!!肥宅大哭orz 题解如下 最后一句:“ ...

  8. 华南理工大学 “三七互娱杯” C HRY and Abaas

    https://ac.nowcoder.com/acm/contest/874/C 题目大意是两人俄罗斯轮盘赌 n个位置 有m个子弹 已知哪些位置上有子弹 子弹打出 游戏结束 求第i次扣动扳机游戏才结 ...

  9. java中重写

    1.重写[针对父类与子类而言]---------即java的多态性[子类与父类间有相同的名称和参数,此方法就被重写Overriding:又称:方法覆盖] 子类对父类的允许访问的方法的实现过程进行重新编 ...

  10. java学习day1

    一.常用的DOS命令 1.打开cmd 窗口键+r --> 输入cmd --> 确认 2.常用的dos命令 dir:列出当前目录下的所有文件及文件夹 md:创建一个新的目录 rd:删除目录 ...