HDU 5842.Lweb and String

407


题目


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



题解



不是最长上升子序列!!!

不是最长上升子序列!!!

不是最长上升子序列!!!

尽管题目一直恶意引导,比赛的时候也有人恶意引导……

字母和数字的映射是自己随便选的,只要能达到最长就行
比如 cba 答案应该是 3

要达到最长,就应该前面的尽可能小,后面的尽可能大
或者说,就是 计数总共出现了多少个不同的字符……


代码


/*
By:OhYee
Github:OhYee
Blog:http://www.oyohyee.com/
Email:oyohyee@oyohyee.com

かしこいかわいい?
エリーチカ!
要写出来Хорошо的代码哦~
*/
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <functional>
#include <bitset>
#include <iomanip> 
using namespace std;

const int maxn = 27;
bool vis[maxn];

void Do() {
    char c = getchar();;

    memset(vis,false,sizeof(vis));
    int ans = 0;
    while(!(c >= 'a'&&c <= 'z'))
        c=getchar();
    while(c >= 'a'&&c <= 'z') {
        int t = c - 'a';
        if(!vis[t]) {
            ans++;
            vis[t] = true;
        }

        c = getchar();
    }
    printf("%d\n",ans);
}

int main() {
    //cin.tie(0);
    //cin.sync_with_stdio(false);
    int T;
    scanf("%d",&T);
    for(int i = 1;i <= T;i++) {
        printf("Case #%d: ",i);
        Do();
    }
    return 0;
}
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