地址:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3961

题目:

ACM (ACMers' Chatting Messenger) is a famous instant messaging software developed by Marjar Technology Company. To attract more users, Edward, the boss of Marjar Company, has recently added a new feature to the software. The new feature can be described as follows:

If two users, A and B, have been sending messages to each other on the last mconsecutive days, the "friendship point" between them will be increased by 1 point.

More formally, if user A sent messages to user B on each day between the (i - m + 1)-th day and the i-th day (both inclusive), and user B also sent messages to user A on each day between the (i - m + 1)-th day and the i-th day (also both inclusive), the "friendship point" between A and B will be increased by 1 at the end of the i-th day.

Given the chatting logs of two users A and B during n consecutive days, what's the number of the friendship points between them at the end of the n-th day (given that the initial friendship point between them is 0)?

Input

There are multiple test cases. The first line of input contains an integer T (1 ≤ T ≤ 10), indicating the number of test cases. For each test case:

The first line contains 4 integers n (1 ≤ n ≤ 109), m (1 ≤ m ≤ n), x and y (1 ≤ xy ≤ 100). The meanings of n and m are described above, while x indicates the number of chatting logs about the messages sent by A to B, and y indicates the number of chatting logs about the messages sent by B to A.

For the following x lines, the i-th line contains 2 integers lai and rai (1 ≤ la,i ≤ rai ≤ n), indicating that A sent messages to B on each day between the lai-th day and the rai-th day (both inclusive).

For the following y lines, the i-th line contains 2 integers lbi and rbi (1 ≤ lb,i ≤ rbi ≤ n), indicating that B sent messages to A on each day between the lbi-th day and the rbi-th day (both inclusive).

It is guaranteed that for all 1 ≤ i < xrai + 1 < lai + 1 and for all 1 ≤ i < yrbi + 1 < lbi + 1.

Output

For each test case, output one line containing one integer, indicating the number of friendship points between A and B at the end of the n-th day.

Sample Input

2
10 3 3 2
1 3
5 8
10 10
1 8
10 10
5 3 1 1
1 2
4 5

Sample Output

3
0

Hint

For the first test case, user A and user B send messages to each other on the 1st, 2nd, 3rd, 5th, 6th, 7th, 8th and 10th day. As m = 3, the friendship points between them will be increased by 1 at the end of the 3rd, 7th and 8th day. So the answer is 3.

思路:手速题+4

暴力n^2匹配就好

 #include <bits/stdc++.h>

 using namespace std;

 #define MP make_pair
#define PB push_back
#define ll first
#define rr second
typedef long long LL;
typedef pair<int,int> PII;
const double eps=1e-;
const double pi=acos(-1.0);
const int K=1e6+;
const int mod=1e9+; int n,m,na,nb;
PII ta[K],tb[K]; int main(void)
{
int t;cin>>t;
while(t--)
{
int ans=;
cin>>n>>m>>na>>nb;
for(int i=;i<=na;i++)
scanf("%d%d",&ta[i].ll,&ta[i].rr);
for(int i=;i<=nb;i++)
scanf("%d%d",&tb[i].ll,&tb[i].rr);
for(int i=;i<=na;i++)
{
for(int j=;j<=nb;j++)
if(tb[j].ll>=ta[i].ll && ta[i].rr>=tb[j].ll)
{
int tmp=min(ta[i].rr,tb[j].rr)-tb[j].ll+;
ans+=tmp>=m?tmp-m+:;
}
else if(tb[j].rr>=ta[i].ll && ta[i].rr>=tb[j].rr)
{
int tmp=tb[j].rr-max(ta[i].ll,tb[j].ll)+;
ans+=tmp>=m?tmp-m+:;
}
else if(ta[i].ll>=tb[j].ll && ta[i].rr<=tb[j].rr)
{
int tmp=ta[i].rr-ta[i].ll+;
ans+=tmp>=m?tmp-m+:;
}
}
printf("%d\n",ans);
}
return ;
}

最新文章

  1. XML技术的应用
  2. 最短JavaScript判断是否为IE6、IE的方法
  3. Scala 深入浅出实战经典 第54讲:Scala中复合类型实战详解
  4. 仅IE6中链接A的href为javascript协议时不能在当前页面跳转
  5. C#计算某一些任务的执行时间(消耗时间)
  6. Java jdk环境变量配置
  7. linux系统的目录讲解
  8. Visual Studio 2015 Update 3 RC 候选预览版粗来了
  9. Delphi中获取Unix时间戳及注意事项(c语言中time()是按格林威治时间计算的,比北京时间多了8小时)
  10. js在方法Ajax请求数据来推断,验证无效(OnClientClick=&amp;quot;return Method();&amp;quot;),或者直接运行的代码隐藏
  11. 前端模块化——seaJS
  12. akka tips
  13. Tengine+Lua+GraphicsMagick
  14. install kali on my x200
  15. 如何把ppt写好
  16. 在hue中使用hive
  17. Linux shell脚本中shift
  18. 一个简易的drf的项目例子
  19. 解决 Sublime text3 中文显示乱码问题【亲测可用】
  20. [javase学习笔记]-8.3 statickeyword使用的注意细节

热门文章

  1. Linux下Apache与httpd的区别与关系
  2. easyui上次图片
  3. ArcGIS -- ArcSDE Lock request conflicts with an established lock
  4. 数据库为什么要用B+树结构--MySQL索引结构的实现(转)
  5. poj_2479 动态规划
  6. input的disable和readonly
  7. docker的私有仓库的搭建
  8. LeetCode 笔记系列五 Generate Parentheses
  9. 利用脚手架vue cli搭建vue项目
  10. Hibernate的二级缓存(SessionFaction的外置缓存)-----Helloword