Lweb and String

Lweb and String

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 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

 

这题用了LIS的nlgn的解法，之后竟然是求字母出现的个数。。。

#include <cstdio>
#include <vector>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=1e5+5;
char s[maxn];
int dp[maxn];
bool vis[30];
int main(){
    //freopen("data.in","r",stdin);
    int t,c=0;
    scanf("%d",&t);
    while(t--){
        scanf("%s",s);
        memset(vis,0,sizeof(vis));
        int len=0;
        int n=strlen(s);
        for(int i=0;i<n;i++)
            vis[s[i]-'a'+0]++;
            /*
        dp[1]=s[0];
        for(int i=1;i<n;i++){
            if(s[i]>dp[len])
                dp[++len]=s[i];
            else {
                int x= lower_bound(dp,dp+len+1,s[i])-dp;
                dp[x]=s[i];
            }
        }
        */
        for(int i=0;i<30;i++)
            if(vis[i]) len++;
        printf("Case #%d: %d\n",++c,len);
    }

}