Jokewithpermutation
Input file: joke.in
Output file: joke.out
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 and output
joke.in
4111109876532
joke.out
4 1 11 10 9 8 7 6 5 3 2
题目链接：http://codeforces.com/gym/100553/attachments/download/2885/20142015-acmicpc-northeastern-european-regional-contest-neerc-14-en.pdf
题意：输入一排数字，所有的数字之间没有空格隔开，这些数字由1-n组，字符串的长度最少为1，最多为50。输出那一排数字，并用空格隔开。
这一题是一道经典的dfs搜索，搜索到的当前状态是字符串的第i个字符，如果这个数字没有被标记，那就搜索dfs(i+1,t+1)；如果i+1<len的话，那么和后面那个字符一起组合成一个数字，如果这个数字不会超过n并且没有被标记的话，那么就搜索dfs(i+2,t+1);如果都不符合的话，那就return回溯。要注意dfs搜索不符合要求的话，要记得把数字的标记状态回溯恢复一下。当i==len时，说明其中一种情况已经搜索完毕也有可能搜索成了一种非正常状态，这时就要用一个for循环遍历判断是否说有的数字均被标记。
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
char s[55];
int flag[55];
int x[55];
int len,n;
int gg;
void dfs(int i,int t);
int main()
{
freopen("joke.in","r",stdin);
freopen("joke.out","w",stdout);
int i;
memset(s,0,sizeof(s));
memset(x,0,sizeof(x));
memset(flag,0,sizeof(flag));
scanf("%s",s);
len=strlen(s);
if(len<10) //优化，如果都是十以内
{
n=len;
for(i=0; i<n-1; i++)
printf("%c ",s[i]);
printf("%c
",s[i]);
}
else
{
n=(len-9)/2+9;
gg=0;
dfs(0,0);
}
return 0;
}
void dfs(int i,int t)
{
if(i==len)
{
int sign=1;
for(int j=1; j<=n; j++)
if(flag[j]==0)
{
sign=0;
break;
}
if(sign)
{
for(int j=0; j<n-1; j++)
printf("%d ",x[j]);
printf("%d
",x[n-1]);
gg=1;
}
return;
}
if(gg==1) return;
if(flag[s[i]-48]==0)
{
flag[s[i]-48]=1;
x[t]=s[i]-48;
dfs(i+1,t+1);
if(gg) return;
flag[s[i]-48]=0;
}
if((i+1)<len&&((s[i]-48)*10+(s[i+1]-48))<=n&&flag[(s[i]-48)*10+(s[i+1]-48)]==0)
{
flag[(s[i]-48)*10+(s[i+1]-48)]=1;
x[t]=(s[i]-48)*10+(s[i+1]-48);
dfs(i+2,t+1);
if(gg) return;
flag[(s[i]-48)*10+(s[i+1]-48)]=0;
}
return;
}