Lweb and String

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 266    Accepted Submission(s): 176 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   解：开始看到LIS以为是递增最大子序列，用dp和二分写完之后，WA，后来才发现，直接统计不同字母数量就可以了。给出LIS对于此题的代码，以及只统计不同字母数量的代码。 dp+二分，n*log(n)： #include<stdio.h> #include<string.h> #include<iostream> #include<algorithm> #include<string.h> using namespace std; char str[100010]; int number[100010]; int dp[100010]; int insert(int num,int end) { int begin=0; while(begin<end-1){ int mid=(begin+end)/2; if(dp[mid]>=num) end=mid; else begin=mid; } dp[begin+1]=num; return begin+1; } int main() { int T; int i,j; int count=0; scanf("%d%*c",&T); while(T--) { memset(number,0,sizeof(number)); count++; gets(str); int leng = strlen(str); for(int i=0;i<100010;i++){ dp[i]=99999; } for(int i=0;i<leng;i++) number[i]=str[i]-'a'+1; sort(number,number+leng); for(int i=0;i<leng;i++){ insert(number[i],leng-1); } printf("Case #%d: %d\n",count,insert(99999,leng)-1); for(int i=0;i<100010;i++){ dp[i]=99999; } } return 0; } 直接统计不同字母数量，n： #include<stdio.h> #include<string.h> int a[27]; char b[100010]; int main() { int t; int count=1; scanf("%d",&t); getchar(); while(t--){ memset(a,0,sizeof(a)); int x=0; gets(b); int len=strlen(b); for(int i=0;i<len;i++){ a[b[i]-'a']=1; } for(int i=0;i<26;i++){ if(a[i]){ x++; } } printf("Case #%d: %d\n",count++,x); } return 0; }