Lweb and String

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 0    Accepted Submission(s): 0 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  

这题说的是映射，即按照出现次序排序，从头到尾都没有说按照字典序排序。。。。。错的都怀疑人生了

#include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <vector> #include <algorithm> using namespace std; const int N = 110000; char str[N]; int a[N], visit[N]; vector<int>G; int binary(int x); int main() {     int ncase=1;     int t;     scanf("%d", &t);     getchar();     while(t--)     {         memset(str,'\0',sizeof(str));         memset(visit,0,sizeof(visit));         gets(str);         int num=1;         for(int i=0;str[i];i++)         {             int x=(str[i]-'a');             if(visit[x]==0)             {                 a[i]=num;                 visit[x]=num;                 num++;             }             else             {                 a[i]=visit[x];             }         }         int len=strlen(str);         G.clear();         for(int i=0;i<len;i++)         {             int tmp=binary(a[i]);             if(tmp>=(int)G.size())             {                 G.push_back(a[i]);             }             else             {                 G[tmp]=a[i];             }         }         int ans=(int) G.size();         printf("Case #%d: %d\n",ncase++,ans);     }     return 0; } int binary(int x) {     int l=0, r=G.size()-1;     while(l<=r)     {         int mid=(l+r)/2;         if(G[mid]<x)         {             l=mid+1;         }         else         {             r=mid-1;         }     }     return l; }