zoj3961（模拟+离散化）

Let’s Chat
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 m consecutive 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.

一开始不知道用什么存储，后来看了一下题解，用的结构体

#include <bits/stdc++.h>
using namespace std;
struct log
{
   
    int a,b,w;
}x[120],y[120];
int main()
{
   
    int t,n,m,p,q,i,j;
    scanf("%d",&t);
    while(t--)
    {
   
        scanf("%d%d%d%d",&n,&m,&p,&q);
        for(i=0;i<p;i++)
        {
   
            scanf("%d%d",&x[i].a,&x[i].b);
            x[i].w=x[i].b-x[i].a+1;
        }
        for(i=0;i<q;i++)
        {
   
            scanf("%d%d",&y[i].a,&y[i].b);
            y[i].w=y[i].b-y[i].a+1;
        }
        int ans=0;
        for(i=0;i<p;i++)
        {
   
            if(x[i].w>=m)
            {
   
                for(j=0;j<q;j++)
                {
   
                    if(y[j].w>=m)
                    {
   
                        int left=max(x[i].a,y[j].a);
                        int right=min(x[i].b,y[j].b);
                        int len=right-left+1;
                        if(len>=m)
                            ans+=len-m+1;
                    }
                }
            }
        }
        cout<<ans<<'\n';
    }
    return 0;
}