POJ 1176 dfs

この問題は本当に気持ちが悪くて、、dfsを書いて提出して、、、wa、よく問題を見て、やっとバイナリの昇順に並べ替えて出力することを発見しました.私は行きます.面倒くさいですね.次は文字列に変換し、文字列で比較し、最後に出力します.の注意すべきは、スイッチを押す回数が4より大きい場合、c%4+4回の効果と同じです.
Party Lamps
Time Limit: 1000MS
 
Memory Limit: 10000K
Total Submissions: 3315
 
Accepted: 1127
Description
To brighten up the gala dinner of the IOI'98 we have a set of N coloured lamps numbered from 
1 to N. The lamps are connected to four buttons: 
button 1 -- when this button is pressed, all the lamps change their state: those that are ON are turned OFF and those that are OFF are turned ON. 
button 2 -- changes the state of all the odd numbered lamps. 
button 3 -- changes the state of all the even numbered lamps. 
button 4 -- changes the state of the lamps whose number is of the form 3K+1 (with K >= 0), i.e., 1,4,7,... 
There is a counter C which records the total number of button presses. 
When the party starts, all the lamps are ON and the counter C is set to zero. 
You are given the value of counter C and information on the final state of some of the lamps. Write a program to determine all the possible final configurations of the N lamps that are consistent with the given information, without repetitions.
Input
Your program is to read from standard input. The input contains four lines, describing the number N of lamps available, the number C of button presses, and the state of some of the lamps in the final configuration. 
The first line contains the number N and the second line the final value of counter C. The third line lists the lamp numbers you are informed to be ON in the final configuration, separated by one space and terminated by the integer -1. The fourth line lists the lamp numbers you are informed to be OFF in the final configuration, separated by one space and terminated by the integer -1. 
The parameters N and C are constrained by: 
10 <= N <= 100 
1 <= C <= 10000 
The number of lamps you are informed to be ON, in the final configuration, is less than or equal to 2.The number of lamps you are informed to be OFF, in the final configuration, is less than or equal to 2.
Output
Your program is to write to standard output. The output must contain all the possible final configurations (without repetitions) of all the lamps. There is at least one possible final configuration. Each possible configuration must be written on a different line. Each line has N characters, where the first character represents the state of lamp 1 and the last character represents the state of lamp N. A 0 (zero) stands for a lamp that is OFF, and a 1 (one) stands for a lamp that is ON. Configurations should be listed in binary ascending order.
Sample Input

10
1
-1
7 -1

Sample Output

0000000000
0101010101
0110110110

acコード:

#include <iostream>
#include <cstdio>
#include <queue>
#include <stdio.h>
#include <string>
#include <algorithm>
#include <string.h>
using namespace std;
int num[110];
int on[2],off[2];
int n,c,k,kk,total;//total          ，k             ，kk             
struct ss{
  char ch[110];
}aa[1010];
void dfs(int x){
	if(x==c+1){
		int flag=1;
		for(int i=0;i<k;++i){
			if(num[on[i]]!=1){
			  flag=0;break;
			}
		}
		for(int i=0;i<kk;++i){
			if(num[off[i]]!=0){
			  flag=0;break;
			}
		}
		if(flag==1){
		  char str[105];
		  int newflag=1;
		  for(int i=1;i<=n;++i)
		  {str[i-1]='0'+num[i];}
		  str[n]='\0';
		  for(int i=0;i<total;++i){
			  if(strcmp(str,aa[i].ch)==0)
			  {flag=0;break;}
		  }
		  if(flag==1){
		  strcpy(aa[total].ch,str);
		  total++;
		  }
		  return;
		}
	}
	else{
		for(int i=1;i<=4;++i){
			if(i==1){
				for(int j=1;j<=n;++j){
				  num[j]=!num[j];
				}
				dfs(x+1);
				for(int j=1;j<=n;++j)
					num[j]=!num[j];
			}
			else if(i==2){
				for(int j=1;j<=n;++j){
				  if(j%2)
					  num[j]=!num[j];
				}
				dfs(x+1);
				for(int j=1;j<=n;++j)
					if(j%2)
						num[j]=!num[j];
			}
			else if(i==3){
				for(int j=1;j<=n;++j){
				  if(j%2==0)
					  num[j]=!num[j];
				}
				dfs(x+1);
				for(int j=1;j<=n;++j)
					if(j%2==0)
						num[j]=!num[j];
			}
			else{
				for(int j=1;j<=n;++j){
				  if((j-1)%3==0)
					  num[j]=!num[j];
				}
				dfs(x+1);
				for(int j=1;j<=n;++j)
					if((j-1)%3==0)
						num[j]=!num[j];
			}
		}
	}
}
int cmp(ss a,ss b){
	return strcmp(a.ch,b.ch)<0;
}
int main(){
	//freopen("11.txt","r",stdin);
	while(~scanf("%d%d",&n,&c)){
		if(c>4){
	      c=c%4+4;
		}
	  int x,y;
	  for(int i=1;i<=n;++i)
	  {num[i]=1;}
	  k=0,kk=0,total=0;
	  while(scanf("%d",&x)&&x!=-1){
	    on[k++]=x;
	  }
	  while(scanf("%d",&y)&&y!=-1)
		 off[kk++]=y;
	  dfs(1);
	  sort(aa,aa+total,cmp);//     
	  for(int i=0;i<total;++i)
		  printf("%s
",aa[i].ch);
	}
  return 0;
}