浙江省赛 D 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 lastmconsecutive 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 thei-th day (both inclusive), and user B also sent messages to user A on each day between the (i -m + 1)-th day and thei-th day (also both inclusive), the "friendship point" between A and B will be increased by 1 at the end of thei-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 then-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, whilex indicates the number of chatting logs about the messages sent by A to B, andy 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 integersl_a,i and r_a,i (1 ≤l_a,i ≤r_a,i ≤n), indicating that A sent messages to B on each day between thel_a,i-th day and the r_a,i-th day (both inclusive).

For the following y lines, the i-th line contains 2 integersl_b,i and r_b,i (1 ≤l_b,i ≤r_b,i ≤n), indicating that B sent messages to A on each day between thel_b,i-th day and the r_b,i-th day (both inclusive).

It is guaranteed that for all 1 ≤ i < x, r_a,i + 1 <l_{a,i + 1} and for all 1 ≤i < y, r_b,i + 1 <l_{b,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 then-th day.

Sample Input

210 3 3 21 35 810 101 810 105 3 1 11 24 5

Sample Output

30

思路：贪心算法：观察可以转化为区间上的公共部分的问题

#include <iostream>#include<cstdio>#include<algorithm>using namespace std;int x[105][105],y[105][105];int main(){    int icase;   // freopen("e:\\in.txt","r",stdin);    scanf("%d",&icase);    while(icase--)    {        int n,m,X,Y;        scanf("%d%d%d%d",&n,&m,&X,&Y);        for(int i=1;i<=X;i++)            for(int j=1;j<=2;j++)              scanf("%d",&x[i][j]);         for(int i=1;i<=Y;i++)            for(int j=1;j<=2;j++)              scanf("%d",&y[i][j]);          int re=0;          for(int i=1;i<=X;i++)          {              if(x[i][2]-x[i][1]<m-1)                 continue;              for(int j=1;j<=Y;j++)              {                  if(y[j][2]-y[j][1]<m-1)                    continue;                  int l=max(y[j][1],x[i][1]),                  r=min(x[i][2],y[j][2]),                  length=r-l+1;                  if(length>=m)                  re+=length-m+1;              }          }          printf("%d\n",re);    }    return 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. Asm = 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.

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