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); } }
