CF716B Complete the Word

题意翻译

给你六个单词，要求拼成一个字典序最小的字符8字形，空的地方用.代替。

题目描述

ZS the Coder loves to read the dictionary. He thinks that a word is nice if there exists a substring (contiguous segment of letters) of it of length 2626 where each letter of English alphabet appears exactly once. In particular, if the string has length strictly less than 2626 , no such substring exists and thus it is not nice.

Now, ZS the Coder tells you a word, where some of its letters are missing as he forgot them. He wants to determine if it is possible to fill in the missing letters so that the resulting word is nice. If it is possible, he needs you to find an example of such a word as well. Can you help him?

输入输出格式

输入格式：

The first and only line of the input contains a single string ss ( 1<=|s|<=500001<=∣s∣<=50000 ), the word that ZS the Coder remembers. Each character of the string is the uppercase letter of English alphabet ('A'-'Z') or is a question mark ('?'), where the question marks denotes the letters that ZS the Coder can't remember.

输出格式：

If there is no way to replace all the question marks with uppercase letters such that the resulting word is nice, then print -1−1 in the only line.

Otherwise, print a string which denotes a possible nice word that ZS the Coder learned. This string should match the string from the input, except for the question marks replaced with uppercase English letters.

If there are multiple solutions, you may print any of them.

输入输出样例

输入样例#1： 复制

ABC??FGHIJK???OPQR?TUVWXY?

输出样例#1： 复制

ABCDEFGHIJKLMNOPQRZTUVWXYS

输入样例#2： 复制

WELCOMETOCODEFORCESROUNDTHREEHUNDREDANDSEVENTYTWO

输出样例#2： 复制

-1

输入样例#3： 复制

??????????????????????????

输出样例#3： 复制

MNBVCXZLKJHGFDSAQPWOEIRUYT

输入样例#4： 复制

AABCDEFGHIJKLMNOPQRSTUVW??M

输出样例#4： 复制

-1


// 暴力。错了很多次，无话可说。
// 字符串中的所有问号都要变，，，，，，#include<iostream>
#include<queue>
#include<cstring>
using namespace std;
char str[50009];
int main(void){while( scanf("%s",str)!=EOF){int len = strlen( str );if( len < 26 ){printf("-1\n");continue;}else {int vis[30];int cnt = 0 , id ,flag = 0;memset(vis,0,sizeof(vis));for( int i=0;i<len;i++){cnt++;id = str[i] - 'A'; if( str[i] != '?' ){if( vis[id] == 0  ){ vis[id] = 1;    }else {i =  i - cnt +1;cnt =0;memset(vis,0,sizeof(vis));}}if( cnt == 26 ){queue<char> q;for( int k = 0;k<=25;k++ ){if(!vis[k]){q.push( 'A'+k);}} for( int j=i;j>=i-25;j--){if( str[j] =='?'){str[j] = q.front();q.pop();      } }for( int l =0;l<len;l++){if(str[l] == '?')str[l] = 'A';} printf("%s\n",str);flag = 1;break; }     } if( !flag ){printf("-1\n");}}}return 0;
}