省赛D Let's Chat

Let’s Chat

Time Limit: 1 Second Memory Limit: 65536 KB

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

这题蛮简单的，题目没有出区间相交很好啊，就是找下区间并集

#include <stdio.h>
#include <algorithm>
using namespace std;
int l[105],r[105];
int main()
{int t;
while(~scanf("%d",&t)){
while(t--){
int n,m,x,y;
scanf("%d%d%d%d",&n,&m,&x,&y);
m-=2;
for(int i=0;i<x;i++)
scanf("%d%d",&l[i],&r[i]);
int s=0;
for(int i=0;i<y;i++){
int b,c;
scanf("%d%d",&b,&c);
for(int j=0;j<x;j++){
if(r[j]<b)
continue;
if(l[j]>c)
break;
int ma=max(l[j],b);
int mi=min(r[j],c);
if(mi-ma-m>0)
s+=mi-ma-m;
}
}
printf("%d\n",s);
}}

return 0;
}