HDU 5842 Lweb and String(Lweb与字符串)

HDU 5842 Lweb and String(Lweb与字符串)

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

 

Problem Description - 题目描述
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)
Lweb有个字符串S

某天,他决定用原串弄腾出一个新串。

你需要帮他确定变化的形式,使得新串具有最长的LIS(严格递增)。

你需要把字符串中的每种字母转化成一个数字。

A是S的字母集合,B是自然数集合。

任意映射 f:A→B 都是合法变动。

比如说,一个字符串“aabc”,A={a,b,c},你可以转换成“1 1 2 3”,新串的LIS长度为3。

现在帮帮Lweb找出你能从S中获得的最长LIS。

LIS:最长升序子序列。(https://en.wikipedia.org/wiki/Longest_increasing_subsequence)
CN
Input - 输入
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.
输入的第一行只有一个整数T,(1≤T≤20)。

随后T行,每行都有一个仅由小写字母组成的字符串S,S的长度不超过10^5
CN

Output - 输出
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.
对于每组测试用例,输出一行"Case #x: y",x表示测试用例的编号,从1开始。y为答案。
CN

Sample Input - 输入样例

2
aabcc
acdeaa

 

Sample Output - 输出样例

Case #1: 3
Case #2: 4

题解

  字母转数字,然后求最长升序子序列的长度。

  因此映射关系是随意自定的,因此题目只要求字符串中的字母种类数量即可。

代码 C++

 1 #include <cstdio>
 2 #include <cstring>
 3 char data[100005], inUS[128];
 4 int main(){
 5     int t, i, j, opt;
 6     scanf("%d ", &t);
 7     for (i = 1; i <= t; ++i){
 8         gets(data);
 9         memset(inUS, 0, sizeof(inUS)); opt = 0;
10         for (j = 0; data[j]; ++j){
11             if (~inUS[data[j]]) ++opt, --inUS[data[j]];
12         }
13         printf("Case #%d: %d
", i, opt);
14     }
15     return 0;
16 }


 

 

原文地址:https://www.cnblogs.com/Simon-X/p/6003442.html