标签：c++   acm   algorithm   dfs   uva   

The House Of Santa Claus

In your childhood you most likely had to solve the riddle of the house of Santa Claus. Do you remember that the importance was on drawing the house in a stretch without lifting the pencil and not drawing a line twice? As a reminder it has to look like shown in Figure 1.

   
Figure: The House of Santa Claus

Well, a couple of years later, like now, you have to ``draw‘‘ the house again but on the computer. As one possibility is not enough, we require all the possibilities when starting in the lower left corner. Follow the example in Figure 2 while defining your stretch.

   
Figure: This Sequence would give the Outputline 153125432

All the possibilities have to be listed in the outputfile by increasing order, meaning that 1234... is listed before 1235... .

Output

So, an outputfile could look like this:

12435123
13245123
...
15123421

题意:  一笔画问题。从1开始画，输出所有可能的情况

题解: DFS 只要用个数组 path保存下路径就可以了~~

AC代码:

#include<iostream>
using namespace std;
bool edge[6][6]={0},visit[6][6];
int path[10];
void dfs(int pos,int step){
	if(step==8){
		for(int i=0;i<9;i++)
		cout<<path[i];
		cout<<endl;
		return ;
	}
	for(int i=1;i<6;i++){
		if(edge[pos][i]&&visit[pos][i]){
			visit[pos][i]=false;
			visit[i][pos]=false;
			path[step+1]=i;
			dfs(i,step+1);
			visit[pos][i]=true;
			visit[i][pos]=true;
		}
		
	}
}
int main(){
	edge[1][2]=true;edge[1][5]=true;edge[1][3]=true;
	edge[2][1]=true;edge[2][5]=true;edge[2][3]=true;
	edge[3][2]=true;edge[3][5]=true;edge[3][4]=true;
	edge[3][1]=true;
	edge[4][3]=true;edge[4][5]=true;edge[5][3]=true;
	edge[5][4]=true;edge[5][1]=true;edge[5][2]=true;
	for(int i=0;i<6;i++)
		for(int j=0;j<6;j++)
			visit[i][j]=edge[i][j];
	path[0]=1;
	dfs(1,0);
	return 0;
}

 

291 - The House Of Santa Claus,布布扣,bubuko.com

291 - The House Of Santa Claus

标签：c++   acm   algorithm   dfs   uva   

原文地址：http://blog.csdn.net/mummyding/article/details/38517413