Lweb and String

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 207    Accepted Submission(s): 131 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 题意：给你一个串求LIS，注意a不一定是最小的，根据他给的顺序。 比如 串是cbdc 那么转换过来是1231

思路：

看错题以为裸LIS，果断WA。

才发现是个更水的题T T

只要判断是否出现过，没出现过+1即可。LIS都可以省了

代码：

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> using namespace std; #define N 100050 char a[N]; int n; int ans[N]; int main() { int T,t,tot=1; scanf("%d",&T); while(T--) { scanf("%s",a); int m=strlen(a); ans[1]=a[0]-'a'; int len=1; for(int i=1; i<m; i++) { if(a[i]-'a'>ans[len]) ans[++len]=a[i]-'a'; else { int pos=lower_bound(ans+1,ans+1+len,a[i]-'a')-ans;///注意这个地方是减ans，长度要从1开始 ans[pos]=a[i]-'a'; } } printf("Case #%d: %d\n",tot++,len); } return 0; }