hdu 4079 Happy Telephones

Happy Telephones

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 199    Accepted Submission(s): 159

Problem Description

In the land of Eden, all phone conversations are happy ones. People complaining on the phone are immediately put in jail. To enforce this law, the police taps all phone conversations. 
The police wants to 
hire the approriate number of operators to listen to all conversations 
in a given period of time. Unfortunately, each of their operators can 
listen to one conversation only before needing a really long break to 
rest from the effort. 
As a contractor of the police department, you 
have been asked to provide a program capable of determining the required
 number of operators. If the program does not work correctly, you will 
be put in jail as well, along with all the unhappy complainers. Do you 
really want to end up there?
 

 

Input

Each test case starts with two integers denoting the number of phone calls N (1 <= N < 10 000) and the number of intervals M (1 <= M < 100). This is followed by N lines describing the telephone calls, each one consisting of four integers Source, Destination, Start and Duration. Source and Destination identify the pair of telephone numbers establishing the connection (0 <= Source, Destination <= 10 000 000). Start and Duration are the start time and duration of the call in seconds (1 <= Duration <= 10 000 and Start >= 0). You can safely assume that the sum of Start and Duration fits into a 32-bit signed integer. 
Afterwards follow M lines containing the time 
intervals the police are interested in, each on described by two 
integers Start and Duration, in the same format and with the same 
meaning and constraints as those in the telephone calls. The last test 
case is represented by N = M = 0 and must not be processed.

 

Output

For each of the M intervals of each test case, print the number of calls that are active during at least one second of the interval.

 

Sample Input

3 2

3 4 2 5

1 2 0 10

6 5 5 8

0 6

8 2

1 2

8 9 0 10

9 1

10 1

0 0

 

Sample Output

3

2

1

0

 

Source

2009 South Western European Regional Contest

 

Recommend

lcy

//很简单的一题、我一开始就想复杂了
//居然用上了线段树、离散化、、、还是错的方法
//其实只要判断2条线段有没有交点
//看到别人提交代码那么短、醒悟过来、、
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#include <cmath>
#define N 10002
using namespace std;
struct node
{
    int x,y;
}st[N];
int main()
{
    int i,j,k;
    int n,m;
    int s,d,f,t;
    while(scanf("%d%d",&n,&m),n||m)
    {
        for(i=0;i<n;i++)
        {
            scanf("%d%d",&s,&d); scanf("%d%d",&f,&t);
            st[i].x=f;st[i].y=f+t;
        }
        for(i=0;i<m;i++)
         {
            scanf("%d%d",&f,&t);t=f+t;
            for(k=0,j=0;j<n;j++)
              {
                  if(st[j].y>f&&st[j].x<t)
                       k++;
              }
            printf("%d\n",k);
         }
    }
    return 0;
}