hdu 5842 Lweb and String

Lweb andString

Time Limit: 2000/1000 MS(Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 4367    Accepted Submission(s): 1265

Problem Description

Lweb has a string S.

Oneday, he decided to transform this string to a new sequence. 

You need help him determine this transformation to get a sequence which has thelongest LIS(Strictly Increasing). 

You need transform every letter in this string to a new number.

A is the set of letters of S, B is the setof natural numbers. 

Every injection f:A→B can be treatas an legal transformation. 

For example, a String “aabc”, A={a,b,c}, and you can transform it to “1 1 2 3”, and the LIS ofthe new sequence is 3. 

Now help Lweb, find the longest LIS which you can obtain from S.

LIS: Longest Increasing Subsequence.(https://en.wikipedia.org/wiki/Longest_increasing_subsequence)

 

 

Input

The first line ofthe input contains the only integer T,(1≤T≤20).

Then T lines follow, the i-th line contains a string S only containing the lowercase letters, the lengthof S will not exceed 105.

 

 

Output

For each testcase, output a single line "Case #x: y", where x is the case number,starting from 1. And y is the answer.

 

 

Sample Input

2

aabcc

acdeaa

 

 

Sample Output

Case #1: 3

Case #2: 4

 

题解：

这一题刚看的时候以为是求最长上升子序列，后来才发现排个序就判断就可以了。

 

代码：

#include <iostream>  
#include <cstdio>  
#include <cstring>  
#include <algorithm>
using namespace std;  
const int N = 100000+5;  
int a[N];       
int d[N];       
char s[N];

int main()  
{  
    int t;  
    int p;  
    scanf("%d",&t);  
    for(int j=0;j<t;j++)  
    {  
    	memset(a,0,sizeof(a));
    	memset(d,0,sizeof(d));
    	memset(s,0,sizeof(s));
    	scanf("%s",s);
        //scanf("%d",&p);  
        p=strlen(s);
        for(int i = 0; i < p; i++)  
        {
            //scanf("%d",&a[i]);  
            a[i]=s[i];
        }
        
        sort(a,a+p);
        
        int len=1;
        
        for(int i=1;i<p;i++)
        {
        	if(a[i]!=a[i-1])
        	{
        		len++;
        	}
        }
        
        printf("Case #%d: %d\n",j+1,len);  
    }  
    return 0;  
}