ZOJ 3961 Let's Chat

Time limit1000 msMemory limit65536 kB

暴力

Description

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：

第一行输入一个t；表示有t组数据；
每一组数据输第一行输入四个数：n,m,x,y;表示一段数轴长为1到n，有两个区间：分别的个数为x和y；问你这两个区间的公共区间有多少个长为m的区间；

解题思路：

直接求两个区间的开头的最大值和结尾的最小值；最终可能的区间个数为min-max-m+2；

完整代码：

#include<bits/stdc++.h>
using namespace std;
const int maxn=110;
struct node
{
    int l,r;
}a[maxn],b[maxn];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,m,x,y;
        int ans=0;
        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);
        for(int i=0;i<y;i++)
            scanf("%d%d",&b[i].l,&b[i].r);
        for(int i=0;i<x;i++)
            for(int j=0;j<y;j++)
            {
                int maxl=max(a[i].l,b[j].l);
                int minr=min(a[i].r,b[j].r);
                if(minr-maxl+1>=m)
                {
                    ans+=minr-maxl-m+2;
                }
            }
    printf("%d\n",ans);
    }
    return 0;
}