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.
题意：给出两人的聊天情况，即第 l 天到第 r 天有给对方发消息（对方不一定会消息），如果两人连续k天互发消息则好感度+1，并且之后互聊的每一天好感度都+1.问两人的好感度最后是多少？
思路：可以看出是一个区间问题，即当两区间有交叉时求两人左天数的最大值和右天数的最小值，然后判断（最小值-最大值）与 k 相比较的大小。最后就很容易得到最后结果了。
1 #include <iostream> 2 #include<cstdio> 3 #include <cstring> 4 #include<algorithm> 5 using namespace std; 6 const int maxn=80005; 7 struct node 8 { 9 int l,r; 10 }; 11 node a[105],b[105]; 12 int cmp1(const node &u,const node &v) 13 { 14 return u.l<v.l; 15 } 16 int main() 17 { 18 int t; 19 while(~scanf("%d",&t)) 20 { 21 while(t--) 22 { 23 int n,m,x,y; 24 scanf("%d%d%d%d",&n,&m,&x,&y); 25 for(int i=0;i<x;i++) 26 { 27 scanf("%d%d",&a[i].l,&a[i].r); 28 } 29 int ans=0,maxx,minn; 30 for(int j=0;j<y;j++) 31 { 32 scanf("%d%d",&b[j].l,&b[j].r); 33 for(int i=0;i<x;i++) 34 { 35 if(a[i].r<b[j].l||b[j].r<a[i].l) 36 continue; 37 maxx=max(a[i].l,b[j].l); 38 minn=min(a[i].r,b[j].r); 39 if(minn-maxx-m+2>0) 40 { 41 ans+=(minn-maxx-m+2); 42 } 43 } 44 } 45 printf("%d ",ans); 46 } 47 } 48 return 0; 49 }