HDU-OJ-1800 Flying to the Mars

Flying to the Mars

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description

In the year 8888, the Earth is ruled by the PPF Empire . As the population growing , PPF needs to find more land for the newborns . Finally , PPF decides to attack Kscinow who ruling the Mars . Here the problem comes! How can the soldiers reach the Mars ? PPF convokes his soldiers and asks for their suggestions . “Rush … ” one soldier answers. “Shut up ! Do I have to remind you that there isn’t any road to the Mars from here!” PPF replies. “Fly !” another answers. PPF smiles :“Clever guy ! Although we haven’t got wings , I can buy some magic broomsticks from HARRY POTTER to help you .” Now , it’s time to learn to fly on a broomstick ! we assume that one soldier has one level number indicating his degree. The soldier who has a higher level could teach the lower , that is to say the former’s level > the latter’s . But the lower can’t teach the higher. One soldier can have only one teacher at most , certainly , having no teacher is also legal. Similarly one soldier can have only one student at most while having no student is also possible. Teacher can teach his student on the same broomstick .Certainly , all the soldier must have practiced on the broomstick before they fly to the Mars! Magic broomstick is expensive !So , can you help PPF to calculate the minimum number of the broomstick needed .
For example : 
There are 5 soldiers (A B C D E)with level numbers : 2 4 5 6 4;
One method :
C could teach B; B could teach A; So , A B C are eligible to study on the same broomstick.
D could teach E;So D E are eligible to study on the same broomstick;
Using this method , we need 2 broomsticks.
Another method:
D could teach A; So A D are eligible to study on the same broomstick.
C could teach B; So B C are eligible to study on the same broomstick.
E with no teacher or student are eligible to study on one broomstick.
Using the method ,we need 3 broomsticks.
……

After checking up all possible method, we found that 2 is the minimum number of broomsticks needed. 
 

Input
Input file contains multiple test cases. 
In a test case,the first line contains a single positive number N indicating the number of soldiers.(0<=N<=3000)
Next N lines :There is only one nonnegative integer on each line , indicating the level number for each soldier.( less than 30 digits);
 

Output
For each case, output the minimum number of broomsticks on a single line.
 

Sample Input
 
   
4 10 20 30 04 5 2 3 4 3 4
 

Sample Output
 
   
1 2
 

Author
PPF@JLU
 

————————————————————吃撑的分割线————————————————————
思路:一开始没有看数据范围,直接用int做的,居然水过了,这说明测试数据根本就不是30位整数吧!就是9位整数好吧……方法很简单,排序,然后取某个重复次数最多的等级,重复次数就是答案。
但是如果数据真的是30位,该怎么排序呢?方法是把字符数组写成一个结构体,然后用结构体进行快排。
代码如下:
/****************************************/
 #include 
 #include 
 #include 
 #include 
 #include 
 #include 
 #include 
 #include 
 #include 
 #include 
 #include 
 using namespace std;
/****************************************/
struct str
{
	char s[32];
}a[3010];

bool cmp(str a, str b)
{
    return strcmp(a.s, b.s) > 0;
}

int main()
{
	int n;
	while(~scanf("%d", &n)) {
		int ans = 1, maxi = 1;
		char ss[32];
		for(int i = 0 ; i < n; i++) {
			scanf("%s", ss);
			int j;
			for(j = 0; j < 32; j++) {
				if(ss[j] != '0')
					break;
			}
			strcpy(a[i].s, ss+j);
		}
		sort(a, a+n, cmp);
		for(int i = 1; i < n; i++) {
			if(!strcmp(a[i].s, a[i-1].s)) {
				ans++;
				maxi = max(maxi, ans);
			}
			else {
				ans = 1;
			}
		}
		printf("%d\n", maxi);
	}
	return 0;
}


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