第14届浙江省赛--Let's Chat

Posted 2020-09-14

Let‘s Chat

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Time Limit: 1 Second      Memory Limit: 65536 KB

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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 l_{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 the l_{a, i}-th day and the r_{a, i}-th day (both inclusive).

For the following y lines, the i-th line contains 2 integers l_{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 the l_{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 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.

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Author: WENG, Caizhi
Source: The 14th Zhejiang Provincial Collegiate Programming Contest Sponsored by TuSimple

题意：给出x,y两种区间，若区间交集长度L>=m，贡献值为L-m+1，求总贡献度。

题解：x，y均<100，直接暴力，时间复杂度O(n^2)。

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 
 4 struct Node
 5 {
 6     int r,l;
 7 }a[100],t;
 8 int main()
 9 {
10     int T;
11     scanf("%d",&T);
12     while(T--)
13     {
14         memset(a,0,sizeof(a));
15         int n,m,x,y,score=0;
16         scanf("%d %d %d %d",&n,&m,&x,&y);
17         for(int i=0;i<x;i++)
18         {
19             scanf("%d %d",&a[i].l,&a[i].r);
20         }
21         for(int i=0;i<y;i++)
22         {
23             scanf("%d %d",&t.l,&t.r);
24             if(t.r-t.l+1<m) continue;
25             for(int j=0;j<x&&a[j].l<t.r;j++)
26             {
27                 if(a[j].r-a[j].l+1<m) continue;
28                 int L,R;
29                 L=max(t.l,a[j].l);
30                 R=min(t.r,a[j].r);
31                 if(R-L+1>=m) score+=R-L+1-m+1;
32             }
33         }
34         printf("%d\n",score);
35     }
36     return 0;
37 }