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): 122    Accepted Submission(s): 83

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
  

 

Author

UESTC

 

Source

2016中国大学生程序设计竞赛 - 网络选拔赛 

 

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//
//  main.cpp
//  Lweb and String
//
//  Created by 张嘉韬 on 16/8/14.
//  Copyright © 2016年 张嘉韬. All rights reserved.
//

#include <iostream>
#include <cstdio>
#include <map>
using namespace std;
const int maxn=100000+10;
int main(int argc, const char * argv[]) {
    //freopen("/Users/zhangjiatao/Documents/暑期训练/input.txt","r",stdin);
    int T;
    scanf("%d",&T);
    for(int t=1;t<=T;t++)
    {
        map<char,int> vis;
        vis.clear();
        int counter=0;
        char str[maxn];
        scanf("%s",str);
        for(int i=0;str[i];i++)
        {
            if(vis[str[i]]==1) continue;
            else
            {
                vis[str[i]]=1;
                counter++;
                if(counter>=26) break;
            }
        }
        printf("Case #%d: %d\n",t,counter);
    }
    return 0;
}