浙江省第十四次ACM程序竞赛 Let's Chat（D题）
ACM (ACMers'Chatting Messenger) is a famous instant messaging software developed by MarjarTechnology Company. To attract more users, Edward, the boss of Marjar Company,has recently added a new feature to the software. The new feature can bedescribed as follows:
If two users, Aand B, have been sending messages to each other on thelast mconsecutive days, the "friendship point"between them will be increased by 1 point.
More formally, ifuser 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 sentmessages 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 endof the i-th day.
Given the chattinglogs of two users A and B during n consecutive days, what'sthe number of the friendship points between them at the end of the n-thday (given that the initial friendship point between them is 0)?
Input
There are multipletest 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 linecontains 4 integers n (1 ≤ n ≤ 109), m (1≤ m ≤ n), x and y (1≤ x, y ≤ 100). The meanings of n and m aredescribed above, while x indicates the number of chatting logsabout the messages sent by A to B, and y indicates the numberof chatting logs about the messages sent by B to A.
For thefollowing x lines, the i-th line contains 2integers 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-thday and the ra, i-th day (bothinclusive).
For thefollowing y lines, the i-th line contains 2integers 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-thday and the rb, i-th day (bothinclusive).
It is guaranteedthat for all 1 ≤ i < x, ra, i +1 < la, i + 1 andfor all 1 ≤ i < y, rb, i +1 < lb, i + 1.
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Output
For each testcase, output one line containing one integer, indicating the number offriendship points between A and B at the end of the n-th day.
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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
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Sample Output
3
0
<h4< dd=""></h4<>
Hint
For the first testcase, 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 pointsbetween them will be increased by 1 at the end of the 3rd, 7th and 8th day. Sothe answer is 3.
求区间的交集！！！
我的代码：
#include <iostream>
#include <cstdio>
#include <map>
using namespace std;
map<int,int> A,B;
int startA[105],startB[105];
int n,m,x,y;
int a,b;
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int point = 0;
scanf("%d%d%d%d",&n,&m,&x,&y);
int k = 0;
for(int i = 0 ; i < x ;i++)
{
scanf("%d%d",&a,&b);
A[a] = b;
startA[k++] = a;
}
k = 0;
for(int i = 0 ; i < y ; i++)
{
scanf("%d%d",&a,&b);
B[a] = b;
startB[k++]= a;
}
for(int i = 0 ; i < x ;i++)//把A的情况都遍历一遍
{
int sa = startA[i];
int ea = A[startA[i]];
if(sa + m-1 <= ea)//如果A的这一段符合条件的话
{
for(int j = 0 ; j < y ; j++)
{
int sb = startB[j];
int eb = B[startB[j]];
int l = sa,r = ea;
if(startB[j]+m-1<= B[startB[j]])
{
l = max(sa,sb);
r = min(ea,eb);
if(r -l+1-m+1>0)
point +=(r - l+1-m+1);
}
}
}
}
printf("%d\n",point);
}
return 0;
}
别人博客简洁的代码：
#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;
}