省赛热身赛15-D - Let's Chat（结构体）

D - Let's Chat

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu

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Description

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.

【分析】

题意：如果连续通话天数大于等于指定天数，友好度point+1；

思路：两个结构体，遍历其中一个进行比较。注意，如果某个时间段小于指定时间，那么这个时间段就不用操作了。

#include<bits/stdc++.h>
using namespace std;
const int maxn=105;
struct node{
	int l,r;
}a[maxn],b[maxn];
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int n,m,x,y;
		scanf("%d%d%d%d",&n,&m,&x,&y);
		for(int i=0;i<x;i++)
			scanf("%d%d",&a[i].l,&a[i].r);
		for(int i=0;i<y;i++)
			scanf("%d%d",&b[i].l,&b[i].r);
		int p=0;
		for(int i=0;i<x;i++)
		{
			if(a[i].r-a[i].l+1<m)continue;
			for(int j=0;j<y;j++)
			{
				if(b[j].r-b[j].l+1<m)continue;
				int left,right;
				right=min(a[i].r,b[j].r);  //注意这里是i和j作比较
				left=max(a[i].l,b[j].l);
				//cout<<"left="<<left<<",right="<<right<<endl;
				int len=right-left+1;//cout<<"len="<<len<<endl;
				if(len>=m)
					p+=len-m+1;
				//cout<<"p="<<p<<endl;
			}
		}
		cout<<p<<endl;
	}
	return 0;
}