Joke with permutation

Joke with permutation

Time Limit: 3000ms, Special Time Limit:7500ms, Memory Limit:65536KB

Total submit users: 85, Accepted users: 57

Problem 13341 : Special judge

Problem description

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!

Input

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.

Output

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.

Sample Input

4111109876532

Sample Output

4 1 11 10 9 8 7 6 5 3 2

题目描述：给你一段数字串，是1~n数字的乱序，要你将其分开输出

解题思路：用字符串数组s存储数字串，暴力搜索，但是这里要注意的是需要求得n通过简单推导n = len>9? (len-2)+9 : 9;

把n作为参数传入dfs(n)，每次在s串中查找n的字符表示用一个标记数组vis标记为n;n=0作为dfs的跳出条件用mark标记。

这里的mark变量相当重要

#include 
#include 
#include 
#include 
using namespace std;
const int maxn = 100001;
int len;
int mark;
int vis[maxn];
char s[maxn];
void dfs(int n){
	if(n==0)
	{
		mark = 1;
	    return ;
	}
	if(n > 9)
	{
		char t1 = n/10+'0';
		char t2 = n%10+'0';
		for(int i = 0;i < len;++ i)
		{
			if(s[i]==t1&&s[i+1]==t2&&!vis[i]&&!vis[i+1])
			{
				vis[i] = vis[i+1]=n;
				dfs(n-1);
				if(mark) return ;	
				vis[i] = vis[i+1]=0;
			}
		}
	}
	else 
	{
		char t = n+'0';
		for(int i = 0;i < len;++ i)
		{
			if(s[i]==t&&!vis[i])
			{
				vis[i] = n;		
				dfs(n-1);
				if(mark) return ;	
				vis[i] = 0;
			}
		}
	}
}
int main()
{
	cin >> s;
	len = strlen(s);
	mark = 0;
	int num = len > 9 ? ((len-9)/2+9) : len;
	// cout << num << endl;
	memset(vis,0,sizeof(vis));
	dfs(num);
	cout << s[0];
	for(int i = 1;i < len;++ i)
	{
		if(vis[i]==vis[i-1])
			cout << s[i];
		else
			cout << " " << s[i];
	}
	cout << endl;
}