hdu 5842 Lweb and String【水题】

Lweb and String

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 118    Accepted Submission(s): 79

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 the longest 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 set of natural numbers. 

Every injection f:A→B can be treat as 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 of the 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 of the 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 length of S will not exceed 105.

 

 

Output

For each test case, 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

 

题目大意：

给你一个字符串，我们可以给予对应字符串每个字符一个合理的值，然后求其最大上升子序列的值。

思路：

既然我们可以自己定义每个字符串中每个字符的值，我们及可以直接定义，每一次遇到一个没出现的字符，output++即可。

Ac代码：

#include<stdio.h>
#include<string.h>
using namespace std;
char a[1515151];
int vis[26];
int main()
{
    int t;
    int kase=0;
    scanf("%d",&t);
    while(t--)
    {
        memset(vis,0,sizeof(vis));
        scanf("%s",a);
        int n=strlen(a);
        int output=0;
        for(int i=0;i<n;i++)
        {
            if(vis[a[i]-'a']==0)vis[a[i]-'a']=1,output++;
        }
        printf("Case #%d: ",++kase);
        printf("%d\n",output);
    }
}