Jokewithpermutation

題目描述

Joey had saved a permutation of integers from 1 to n in a text file.
All the numbers were written as decimal numbers without leading
spaces. Then Joe made a practical joke on her: he removed all the
spaces in the file. Help Joey to restore the original permutation
after the Joe’s joke! 輸入描述: The input file contains a single line with
a single string — the Joey’s permutation without spaces. The Joey’s
permutation had at least 1 and at most 50 numbers. 輸出描述: Write a line
to the output file with the restored permutation. Don’t forget the
spaces! If there are several possible original permutations, write any
one of them.

示例1
輸入
復制

4111109876532

輸出
復制

4 1 11 10 9 8 7 6 5 3 2

備注:
Author: Mikhail Dvorkin

題意：

給你一串數(shù)，這串數(shù)字為n的全排列，問怎么將數(shù)分段格，使得成為n的全排列
看樣例4111109876532，可以為4 1 11 10 9 8 7 6 5 3 2

題解：

隊友做得，dfs暴力即可，就是假設當為一位數(shù)或者兩位數(shù)時，看看情況
詳細看代碼吧

代碼：

#include<bits/stdc++.h> #define maxn 6000 using namespace std; char a[maxn]; vector<int>ans; bool vis[maxn]; unordered_map<int,int>p; bool f=0; int s; void dfs(int n){if(f)return ;int size=strlen(a+1);if(n>size){printf("%d",ans[0]);for(int i=1;i<ans.size();i++){printf(" %d",ans[i]);}f=1;return ;}if(p[a[n]-'0']==0){p[a[n]-'0']=1;ans.push_back(a[n]-'0');dfs(n+1);ans.pop_back();p[a[n]-'0']=0;}if(f)return ;if(n<=size-1&&p[(a[n]-'0')*10+(a[n+1]-'0')]==0&&(a[n]-'0')*10+(a[n+1]-'0')<=s){p[(a[n]-'0')*10+(a[n+1]-'0')]=1;ans.push_back((a[n]-'0')*10+(a[n+1]-'0'));dfs(n+2);ans.pop_back();p[(a[n]-'0')*10+(a[n+1]-'0')]=0;}if(f)return ; } int main(){scanf("%s",a+1);s=strlen(a+1)>9?(strlen(a+1)-9)/2+9:strlen(a+1);dfs(1);return 0; }

總結

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