Joke with permutation 分类： ACM 2015-08-03 14:09 1

Joke with permutation
Time Limit: 3000ms, Special Time Limit:7500ms, Memory Limit:65536KB
Total submit users: 87, Accepted users: 60
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

Problem Source
NEERC 2014
题意是给出一个全排列的字符串，然后让你确定排列两个元素之间的空格位置。
先根据字符串的长度求出n是多大，然后用dfs搜索符合条件的排列。

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#define maxn 500
using namespace std;
int n,l,cnt;
int flag[maxn];
char str[maxn];
int hash[maxn];
bool ll; 
void dfs(int k)//k表示当前处理到哪个字符
{
        if(ll)
        return ;
        if(k>=n)
        {
        for(int i=0;i<n;i++)
        {
            printf("%c",str[i]) ;
            if(flag[i]&&i!=n-1)
            printf(" ");  

        } //当K>n时，输出排列，return 
        ll=1; 
        return ;
         }
        if(!hash[str[k]-'0']&&(k+1>=n||str[k+1]!='0'))//一个字有两种组合，要么自己组成一个数，要么和后面的字符组成两位数，注意组合完之后, 字符串的第一个不能是零
        {
            hash[str[k]-'0']=1;
            flag[k]=1;
            dfs(k+1);
            flag[k]=0;//整数不可取，回退
            hash[str[k]-'0']=0;
        }
       if(!hash[(str[k]-'0')*10+str[k+1]-'0']&&((str[k]-'0')*10+(str[k+1]-'0'))<=cnt&&(str[k+2]!='0'||k+2>=n))    
       {
              hash[(str[k]-'0')*10+str[k+1]-'0']=1;
           flag[k+1]=1;
           dfs(k+2);
           flag[k+1]=0;
           hash[(str[k]-'0')*10+str[k+1]-'0']=0;
    }           
    }
int main()
{
    while(scanf("%s",str)!=EOF)
    {
        n=strlen(str);
        if(n<=9)
        {
            for(int i=0;i<n-1;i++)
            printf("%c ",str[i]);
            printf("%c\n",str[n-1]); 
            continue; 
        }    
        int num=n-9;
        cnt=9+num/2;
        ll=0; 
        memset(hash,0,sizeof(hash));
        memset(flag,0,sizeof(flag));
        dfs(0);
        printf("\n");
    }
    return 0;
}

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