地址：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3961
题目：
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 mconsecutive 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.
思路：手速题+4
暴力n^2匹配就好
1 #include <bits/stdc++.h>
2
3 using namespace std;
4
5 #define MP make_pair
6 #define PB push_back
7 #define ll first
8 #define rr second
9 typedef long long LL;
10 typedef pair<int,int> PII;
11 const double eps=1e-8;
12 const double pi=acos(-1.0);
13 const int K=1e6+7;
14 const int mod=1e9+7;
15
16 int n,m,na,nb;
17 PII ta[K],tb[K];
18
19
20 int main(void)
21 {
22 int t;cin>>t;
23 while(t--)
24 {
25 int ans=0;
26 cin>>n>>m>>na>>nb;
27 for(int i=1;i<=na;i++)
28 scanf("%d%d",&ta[i].ll,&ta[i].rr);
29 for(int i=1;i<=nb;i++)
30 scanf("%d%d",&tb[i].ll,&tb[i].rr);
31 for(int i=1;i<=na;i++)
32 {
33 for(int j=1;j<=nb;j++)
34 if(tb[j].ll>=ta[i].ll && ta[i].rr>=tb[j].ll)
35 {
36 int tmp=min(ta[i].rr,tb[j].rr)-tb[j].ll+1;
37 ans+=tmp>=m?tmp-m+1:0;
38 }
39 else if(tb[j].rr>=ta[i].ll && ta[i].rr>=tb[j].rr)
40 {
41 int tmp=tb[j].rr-max(ta[i].ll,tb[j].ll)+1;
42 ans+=tmp>=m?tmp-m+1:0;
43 }
44 else if(ta[i].ll>=tb[j].ll && ta[i].rr<=tb[j].rr)
45 {
46 int tmp=ta[i].rr-ta[i].ll+1;
47 ans+=tmp>=m?tmp-m+1:0;
48 }
49 }
50 printf("%d
",ans);
51 }
52 return 0;
53 }