2017浙江省赛 D - Let's Chat ZOJ - 3961

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 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.

思路：手速题+4

暴力n^2匹配就好

#include <bits/stdc++.h>

using namespace std;

#define MP make_pair
#define PB push_back
#define ll first
#define rr second
typedef long long LL;
typedef pair<int,int> PII;
const double eps=1e-8;
const double pi=acos(-1.0);
const int K=1e6+7;
const int mod=1e9+7;

int n,m,na,nb;
PII ta[K],tb[K];


int main(void)
{
    int t;cin>>t;
    while(t--)
    {
        int ans=0;
        cin>>n>>m>>na>>nb;
        for(int i=1;i<=na;i++)
            scanf("%d%d",&ta[i].ll,&ta[i].rr);
        for(int i=1;i<=nb;i++)
            scanf("%d%d",&tb[i].ll,&tb[i].rr);
        for(int i=1;i<=na;i++)
        {
            for(int j=1;j<=nb;j++)
            if(tb[j].ll>=ta[i].ll && ta[i].rr>=tb[j].ll)
            {
                int tmp=min(ta[i].rr,tb[j].rr)-tb[j].ll+1;
                ans+=tmp>=m?tmp-m+1:0;
            }
            else if(tb[j].rr>=ta[i].ll && ta[i].rr>=tb[j].rr)
            {
                int tmp=tb[j].rr-max(ta[i].ll,tb[j].ll)+1;
                ans+=tmp>=m?tmp-m+1:0;
            }
            else if(ta[i].ll>=tb[j].ll && ta[i].rr<=tb[j].rr)
            {
                int tmp=ta[i].rr-ta[i].ll+1;
                ans+=tmp>=m?tmp-m+1:0;
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}