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;}