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.

题意:

给出x(1<=x<=100)个区间和y(1<=y<=100)个区间,求出存在几个长度为m(1<=m<=n)公共子区间。

把题目样例看懂了,基本上题目就会做了。

// Asimple
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <queue>
#include <vector>
#include <string>
#include <cstring>
#include <stack>
#define INF 0x3f3f3f3f
#define mod 2016
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = ;
int n, m, T, len, cnt, num, Max;
int x, y; struct node{
int l;
int r;
};
node a[maxn], b[maxn]; void input() {
scanf("%d", &T);
while( T -- ) {
cin >> n >> m >> x >> y;
memset(a, , sizeof(a));
memset(b, , sizeof(b));
for(int i=; i<x; i++) {
cin >> a[i].l >> a[i].r;
}
for(int i=; i<y; i++) {
cin >> b[i].l >> b[i].r;
}
int cnt = ;
for(int i=; i<x; i++) {
if( a[i].r-a[i].l+ < m ) continue;
for(int j=; j<y; j++) {
if( b[j].r-b[j].l+ < m ) continue;
int l = max(a[i].l, b[j].l);
int r = min(a[i].r, b[j].r);
if( r - l + >= m ) {
cnt += r-l+ -m+;
}
}
}
cout << cnt << endl;
}
} int main() {
input();
return ;
}

Let's Chat ZOJ - 3961的更多相关文章

  1. 2017浙江省赛 D - Let's Chat ZOJ - 3961

    地址:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3961 题目: ACM (ACMers' Chatting Messe ...

  2. ZOJ 3961 Let's Chat 【水】

    题目链接 http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3961 题意 给出两个人的发消息的记录,然后 如果有两人在连续M天 ...

  3. ZOJ - 3961 Let's Chat(区间相交)

    题意:给定一个长度为n的序列,A和B两人分别给定一些按递增顺序排列的区间,区间个数分别为x和y,问被A和B同时给定的区间中长度为m的子区间个数. 分析: 1.1 ≤ n ≤ 109,而1 ≤x, y  ...

  4. ZOJ People Counting

    第十三届浙江省大学生程序设计竞赛 I 题, 一道模拟题. ZOJ  3944http://www.icpc.moe/onlinejudge/showProblem.do?problemCode=394 ...

  5. ZOJ 3686 A Simple Tree Problem

    A Simple Tree Problem Time Limit: 3 Seconds      Memory Limit: 65536 KB Given a rooted tree, each no ...

  6. 三周,用长轮询实现Chat并迁移到Azure测试

    公司的OA从零开始进行开发,继简单的单点登陆.角色与权限.消息中间件之后,轮到在线即时通信的模块需要我独立去完成.这三周除了逛网店见爱*看动漫接兼职,基本上都花在这上面了.简单地说就是用MVC4基于长 ...

  7. Socket programing(make a chat software) summary 1:How to accsess LAN from WAN

    First we should know some basic conceptions about network: 1.Every PC is supposed to have its own IP ...

  8. ZOJ Problem Set - 1394 Polar Explorer

    这道题目还是简单的,但是自己WA了好几次,总结下: 1.对输入的总结,加上上次ZOJ Problem Set - 1334 Basically Speaking ac代码及总结这道题目的总结 题目要求 ...

  9. ZOJ Problem Set - 1392 The Hardest Problem Ever

    放了一个长长的暑假,可能是这辈子最后一个这么长的暑假了吧,呵呵...今天来实验室了,先找了zoj上面简单的题目练练手直接贴代码了,不解释,就是一道简单的密文转换问题: #include <std ...

随机推荐

  1. sap 类的左侧导航栏

  2. Py中enumerate方法【转载】

    转自:http://www.runoob.com/python/python-func-enumerate.html enumerate(sequence, [start=0]) sequence - ...

  3. 安装ElasticSearch5.5.2 注意事项

    官方文档中建议生产环境中打开 bootstrap.memory_lock: true 打开之后会报很多错误要优化一下系统参数 vim /etc/security/limits.conf * soft ...

  4. 【LeetCode每天一题】String to Integer (atoi)(字符串转换成数字)

    Implement atoi which converts a string to an integer.The function first discards as many whitespace ...

  5. VS Code的golang开发配置 之 代码提示

    之前用VS Code的时候,发现自己的代码的提示一直不好,换用JetBrain的Goland的代码提示是好了,但是比较占用资源.在网上找了一些资料,发现很多人也是遇到第三方或者自己的代码无法提示的情况 ...

  6. Java Selenium - 元素操作 (四)

    四,弹出框 京东购物车为例 , 点击‘删除’ 或者‘移到我的关注’ ,就会弹出下面这个框框,练吧: (其实这也不是常规的弹出框,二是div css前端技术做的效果,本想做个Alert的案例,实在不好找 ...

  7. WCF&AppFabric :异常消息: 内存入口检查失败

    bug描述 发件人信息: System.ServiceModel.ServiceHostingEnvironment+HostingManager/31242459 异常: System.Servic ...

  8. js判断当前页面是否有父页面,页面部分跳转解决办法,子页面跳转父页面不跳转解决 (原)

    //如果当前页面存在父页面,则当前页面的父页面重新加载(即子页面父页面连带跳转) if(top.location!=self.location){         window.parent.loca ...

  9. Tomcat 9 和tomcat 8区别以及 tomcat9 新特性

    1.Tomcat 9.0.0.M1 (alpha) 版本的主要特点 详细信息请点击:Tomcat 9.0.0.M1 其他版本信息:详细参见官网,传送门 注明:当前版本要求最低的Java环境为 1.8+ ...

  10. python读取大文件

    最近在学习python的过程中接触到了python对文件的读取.python读取文件一般情况是利用open()函数以及read()函数来完成: f = open(filename,'r') f.rea ...