AtCoder Beginner Contest 089 D - Practical Skill Test
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 |x−i|+|y−j| 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 Ri−Li is a multiple of D.
For each test, find the sum of magic points consumed during that test.
Constraints
- 1≤H,W≤300
- 1≤D≤H×W
- 1≤Ai,j≤H×W
- Ai,j≠Ax,y((i,j)≠(x,y))
- 1≤Q≤105
- 1≤Li≤Ri≤H×W
- (Ri−Li) 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
- AtCoder Beginner Contest 089完整题解
A - Grouping 2 Time limit : 2sec / Memory limit : 256MB Score : 100 points Problem Statement There a ...
- AtCoder Beginner Contest 100 2018/06/16
A - Happy Birthday! Time limit : 2sec / Memory limit : 1000MB Score: 100 points Problem Statement E8 ...
- 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 ...
- AtCoder Beginner Contest 053 ABCD题
A - ABC/ARC Time limit : 2sec / Memory limit : 256MB Score : 100 points Problem Statement Smeke has ...
- 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 ...
- 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\(g(x)=(x-i)^{P-1}\) ...
- AtCoder Beginner Contest 076
A - Rating Goal Time limit : 2sec / Memory limit : 256MB Score : 100 points Problem Statement Takaha ...
- 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& ...
- 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
- org.hibernate.MappingException: An AnnotationConfiguration instance is required to use <mapping clas
在Hibernate中使用annotation出现错误.如题目所示. HibernateSessionFactory类中: // private static Configuration co ...
- vim系统剪切板
原文地址 1.vim常用复制粘贴命令 Vim的复制粘贴命令无疑是y (yank),p(paster),加上yy,P PS: vim有个很有意思的约定(我觉得是一种约定),就是某个命令的大小写都是实现某 ...
- 七:flask-一些小细节
1.在局域网中,让其他电脑访问我的网站:host参数 如果设置为0.0.0.0,则在局域网中,输入当前项目所在的ip+端口就可以访问这个项目如果host设置为固定的ip,如host=‘'192.168 ...
- Selenium学习之==>Xpath使用方法
一.什么是Xpath XPath是XML的路径语言,通俗一点讲就是通过元素的路径来查找到这个标签元素. 工具 Xpath的练习建议大家安装火狐浏览器后,下载插件,FireBug.由于最新版火狐不再支持 ...
- php7.2 下安装yaf扩展
wget http://pecl.php.net/get/yaf-3.0.7.tgz 解压并进入目录: 1 tar -zxvf yaf-3.0.7* && cd yaf-3.0.7 ...
- vue组件之间通信总结---点赞
总结:父组件-->子组件 ①通过属性 步骤1: <son myName="michael" myPhone='123'></son> <son ...
- postfix无法启动问题
open /etc/postfix/main.cf comment out inet_interfaces: all add inet_protocol: ipv4
- Python 入门之 Python三大器 之 迭代器
Python 入门之 Python三大器 之 迭代器 1.迭代器 (1)可迭代对象: <1> 只要具有__ iter __()方法就是一个可迭代对象 (我们可以通过dir()方法去判断一个 ...
- jquery中的插件EChars的使用
首先,进入EChars的官网下载页面:http://echarts.baidu.com/download.html 下载自己需要的版本. 引入jquery包和echars,进入官网的实例:htt ...
- wordpress开发的一些积累
wordpress 攒知识点 记录开发 wordpress 的一些技能点,以备不时之需 短代码 Shortcode 虽然很多插件都是提供,直接在代码中插入类似[Shortcode] 便可以生效,但是很 ...