浙江省第十四次ACM程序竞赛 Let's Chat（D题）

痛定思痛。 2024-02-17 21:26 14阅读 0赞

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 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

题意：A、B两人互发消息，若连续互发m天，这友谊指数+1，求最终的友谊指数，如n=5，m=3，两人在1-4天都互发消息，则友谊为2，即 1 2 3 ，2 3 4 两个连续的三天。

分析：找到他们的交集，然后判断即可。

AC代码：

#include<cstdio>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    const int maxn=100+10;
    struct node{
    	int l,r;
    }X[maxn],Y[maxn];
    int n,m,x,y;
    
    int main(){
    	int T;
    	scanf("%d",&T);
    	while(T--){
    		scanf("%d%d%d%d",&n,&m,&x,&y);
    		for(int i=1;i<=x;i++)
    		scanf("%d%d",&X[i].l,&X[i].r);
    		for(int i=1;i<=y;i++)
    		scanf("%d%d",&Y[i].l,&Y[i].r);
    		
    		int tol=0;
    		for(int i=1;i<=x;i++)
    		{
    			if(X[i].r-X[i].l+1 < m)continue;  //小于m就不用进行下面操作了 
    			
    			for(int j=1;j<=y;j++)
    			{
    				if(Y[j].r-Y[j].l+1 < m)continue;
    				
    				int L,R;
    				R=min(X[i].r,Y[j].r);
    				L=max(X[i].l,Y[j].l);
    				int len=R-L+1;
    				if(len>=m){
    					tol+=(len-m+1);
    				}
    			}
    		}
    		printf("%d\n",tol);
    	}
    	return 0;
    }