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 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 ≤ x, y ≤ 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 la, i and ra, i (1 ≤ la,i ≤ ra, i ≤ n), indicating that A sent messages to B on each day between the la, i-th day and the ra, i-th day (both inclusive).
For the following y lines, the i-th line contains 2 integers lb, i and rb, i (1 ≤ lb,i ≤ rb, i ≤ n), indicating that B sent messages to A on each day between the lb, i-th day and the rb, i-th day (both inclusive).
It is guaranteed that for all 1 ≤ i < x, ra, i + 1 < la, i + 1 and for all 1 ≤ i < y, rb, i + 1 < lb, i + 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
- XML技术的应用
- 最短JavaScript判断是否为IE6、IE的方法
- Scala 深入浅出实战经典 第54讲:Scala中复合类型实战详解
- 仅IE6中链接A的href为javascript协议时不能在当前页面跳转
- C#计算某一些任务的执行时间(消耗时间)
- Java jdk环境变量配置
- linux系统的目录讲解
- Visual Studio 2015 Update 3 RC 候选预览版粗来了
- 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)
- js在方法Ajax请求数据来推断,验证无效(OnClientClick=&;quot;return Method();&;quot;),或者直接运行的代码隐藏
- 前端模块化——seaJS
- akka tips
- Tengine+Lua+GraphicsMagick
- install kali on my x200
- 如何把ppt写好
- 在hue中使用hive
- Linux shell脚本中shift
- 一个简易的drf的项目例子
- 解决 Sublime text3 中文显示乱码问题【亲测可用】
- [javase学习笔记]-8.3 statickeyword使用的注意细节
热门文章
- Linux下Apache与httpd的区别与关系
- easyui上次图片
- ArcGIS -- ArcSDE Lock request conflicts with an established lock
- 数据库为什么要用B+树结构--MySQL索引结构的实现(转)
- poj_2479 动态规划
- input的disable和readonly
- docker的私有仓库的搭建
- LeetCode 笔记系列五 Generate Parentheses
- 利用脚手架vue cli搭建vue项目
- Hibernate的二级缓存(SessionFaction的外置缓存)-----Helloword