浙江省赛--D

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.

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。

#include <iostream>
#include<cstdio>
#include <cstring>
#include<algorithm>
using namespace std;
const int maxn=80005;
struct node
{
    int l,r;
};
node a[105],b[105];
int cmp1(const node &u,const node &v)
{
    return u.l<v.l;
}
int main()
{
    int t;
    while(~scanf("%d",&t))
    {
        while(t--)
        {
            int n,m,x,y;
            scanf("%d%d%d%d",&n,&m,&x,&y);
            for(int i=0;i<x;i++)
            {
                scanf("%d%d",&a[i].l,&a[i].r);
            }
            int ans=0,maxx,minn;
            for(int j=0;j<y;j++)
            {
                scanf("%d%d",&b[j].l,&b[j].r);
                for(int i=0;i<x;i++)
                {
                    if(a[i].r<b[j].l||b[j].r<a[i].l)
                        continue;
                    maxx=max(a[i].l,b[j].l);
                    minn=min(a[i].r,b[j].r);
                    if(minn-maxx-m+2>0)
                    {
                        ans+=(minn-maxx-m+2);
                    }
                }
            }
            printf("%d\n",ans);
        }
    }
    return 0;
}

 

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