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

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.

Thinking

先將a輸入，接著邊輸入b邊遍歷a 看看有沒有滿足條件的，如果滿足則加友情點。 

注意Hint裡說的情況，可以找出算友情點的規律喔~

code

#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <map> 

using namespace std;

class Node {
    public:
        int left;
        int right;
        Node() {
            left = -1;
            right = -1;
        }
};

int main()  
{
    int tN;
    cin >> tN;
    for (int i = 0; i < tN; i++) {
        int n, m;
        cin >> n >> m;
        int aN, bN;
        cin >> aN >> bN;
        vector<Node> vec;
        for (int j = 0; j < aN; j++) {
            Node no;
            cin >> no.left >> no.right;
            if (no.right > n)  {
                no.right = n;
            }
            vec.push_back(no);
        }
        int num = 0;
        for (int j = 0; j < bN; j++) {
            int l, r;
            cin >> l >> r;
            if (r > n)  {
                r = n;
            }
            vector<Node>::iterator it = vec.begin();
            while (it != vec.end()) {
                if (l <= it->left) {
                    if (it->right <= r) {
                        int temp = it->right - it->left - m + 2;
                        if (temp > 0) {
                            num += temp;
                        }
                    } else {
                        int temp = r - it->left - m + 2;
                        if (temp > 0) {
                            num += temp;
                        }
                    }
                } else {
                    if (it->right >= r) {
                        int temp = r - l - m + 2;
                        if (temp > 0) {
                            num += temp;
                        }
                        it++;
                    } else {
                        int temp = it->right - l - m + 2;
                        if (temp > 0) {
                            num += temp;
                        }
                    }
                }
                it++;
            }
        }
        cout << num << endl;
    }
    return 0;  
}