HDU 3650 Hot Expo

Hot Expo

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 886    Accepted Submission(s): 423

Problem Description

Sunny wants to go to the Shanghai Expo this month and he intends to visit n (1 <= n <= 100 ) country pavilions. As we all know, in each pavilion, there is always a wonderful performance every day. Every performance will always be played only one time each day. And performance i begins at beg[i] second and ends at end[i] second (0<=beg[i], end[i] <24 * 3600). Sunny can not visit more than one country pavilion at the same time. Sunny wouldn't like to miss any performance in the country pavilions that he plans to visit. However, it's also well known that getting accommodation in Shanghai is a little expensive, so he has to make a good arrangement to make his staying time (in days) there minimum.

 

Input

The input contains several test cases. Each test case begins with an integer number n. Then n lines follow and each contains two integers representing the begin time and end time of the performance in the corresponding day. 
Input is terminated by a value of zero (0) for n.

 

Output

Output the minimum days that Sunny should spend in visiting in a seperated line for each test case.

 

Sample Input

  

   2
1 4
4 5
2
2 3
4 6
0
  

 

Sample Output

  

   2 
1
  

 

Source

2010 Asia Regional Hangzhou Site —— Online Contest

 

最多只有八万多个点，直接暴力点就行，看重复次数最多的点。

代码如下：

#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<ctime>

using namespace std;

int a[102], b[102];
int ans[86402];
int main()
{
#ifdef test
    freopen("in.txt", "r", stdin);
#endif
    int n;
    while(scanf("%d", &n), n)
    {
        int start = 86402, end = -1;
        memset(ans, 0, sizeof(ans));
        for(int i = 0; i < n; i++)
        {
            scanf("%d%d", &a[i], &b[i]);
            for(int j = a[i]; j <= b[i]; j++)
                ans[j]++;
            if(a[i] < start)
                start = a[i];
            if(b[i] > end)
                end = b[i];
        }
        int max = 0;
        for(int i = start; i < end; i++)
            if(max < ans[i])
                max = ans[i];
        printf("%d\n", max);
    }
    return 0;
}