poj3083解题报告

Children of the Candy Corn

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4046		Accepted: 1838

Description

The cornfield maze is a popular Halloween treat. Visitors are shown the entrance and must wander through the maze facing zombies, chainsaw-wielding psychopaths, hippies, and other terrors on their quest to find the exit.

One popular maze-walking strategy guarantees that the visitor will eventually find the exit. Simply choose either the right or left wall, and follow it. Of course, there's no guarantee which strategy (left or right) will be better, and the path taken is seldom the most efficient. (It also doesn't work on mazes with exits that are not on the edge; those types of mazes are not represented in this problem.)

As the proprieter of a cornfield that is about to be converted into a maze, you'd like to have a computer program that can determine the left and right-hand paths along with the shortest path so that you can figure out which layout has the best chance of confounding visitors.

Input

Input to this problem will begin with a line containing a single integer n indicating the number of mazes. Each maze will consist of one line with a width, w, and height, h (3 <= w, h <= 40), followed by h lines of w characters each that represent the maze layout. Walls are represented by hash marks ('#'), empty space by periods ('.'), the start by an 'S' and the exit by an 'E'.

Exactly one 'S' and one 'E' will be present in the maze, and they will always be located along one of the maze edges and never in a corner. The maze will be fully enclosed by walls ('#'), with the only openings being the 'S' and 'E'. The 'S' and 'E' will also be separated by at least one wall ('#').

You may assume that the maze exit is always reachable from the start point.

Output

For each maze in the input, output on a single line the number of (not necessarily unique) squares that a person would visit (including the 'S' and 'E') for (in order) the left, right, and shortest paths, separated by a single space each. Movement from one square to another is only allowed in the horizontal or vertical direction; movement along the diagonals is not allowed.

Sample Input

2
8 8
########
#......#
#.####.#
#.####.#
#.####.#
#.####.#
#...#..#
#S#E####
9 5
#########
#.#.#.#.#
S.......E
#.#.#.#.#
#########

Sample Output

37 5 5
17 17 9

题目大意:给定一个迷宫，S是起点，E是终点，#是墙不可走，.可以走。先输出贴左边墙走到E的步数，再输出贴

右边墙的，再输出最少走多少部能到E。。。

思路：这个是个综合题，BFS和DFS全用上了，代码快3KB了，我靠的，少考虑了一个情况，WA了许久，终于A了 我靠的。。。

8 8
########
#....#.#
#....#.#
#....#.#
#....#.#
#....#.#
S......#
#E######

少考虑了S转个弯就是E的情况。。WA了许久，终于发现这个BUG了。。。。汗。。

#include<iostream>
using namespace std;
char map[41][41],lock[41][41];	int head,rear,m,n,queue[2000];
struct mm
{
	int x,y;
}mm[4]={-1,0,0,1,1,0,0,-1};
void push(int i){queue[rear++%2000]=i;}
int pop(){return queue[head++%2000];}

int bfs()
{
	int x,y,step,i;
	while(head!=rear)
	{
		x=pop();	y=pop();	step=pop();
		if(map[x][y]=='E')	return step;
		for(i=0;i<4;i++)
			if(x+mm[i].x>-1&&x+mm[i].x<m&&y+mm[i].y>-1&&y+mm[i].y<n&&(map[x+mm[i].x][y+mm[i].y]=='.'||map[x+mm[i].x][y+mm[i].y]=='E')&&lock[x+mm[i].x][y+mm[i].y]==0)
			{	push(x+mm[i].x);	push(y+mm[i].y);	push(step+1);	lock[x+mm[i].x][y+mm[i].y]=1;	}
	}
	return 0;
}

int ldfs(int x,int y,int f)
{
	if(map[x][y]=='E')
		return 1;
	if((map[x+mm[f].x][y+mm[f].y]=='.'||map[x+mm[f].x][y+mm[f].y]=='E')&&map[x+mm[(f+3)%4].x][y+mm[(f+3)%4].y]=='#')
		return 1+ldfs(x+mm[f].x,y+mm[f].y,f);
	else
		if(!(map[x+mm[(f+3)%4].x][y+mm[(f+3)%4].y]=='#'))
		{
			return 1+ldfs(x+mm[(f+3)%4].x,y+mm[(f+3)%4].y,(f+3)%4);
		}
		else
			return ldfs(x,y,(f+1)%4);
}

int rdfs(int x,int y,int f)
{
	if(map[x][y]=='E')
		return 1;
	if((map[x+mm[f].x][y+mm[f].y]=='.'||map[x+mm[f].x][y+mm[f].y]=='E')&&map[x+mm[(f+1)%4].x][y+mm[(f+1)%4].y]=='#')
		return 1+rdfs(x+mm[f].x,y+mm[f].y,f);
	else
		if(!(map[x+mm[(f+1)%4].x][y+mm[(f+1)%4].y]=='#'))
		{
			return 1+rdfs(x+mm[(f+1)%4].x,y+mm[(f+1)%4].y,(f+1)%4);
		}
		else
			return rdfs(x,y,(f+3)%4);
}


int main()
{
	int t,i,j,x,y,f,o;
	cin>>t;
	while(t--)
	{
		memset(lock,0,sizeof(lock));
		head=rear=0;
		cin>>n>>m;
		for(i=0;i<m;i++)
			cin>>map[i];
		for(i=0;i<m;i++)
			for(j=0;j<n;j++)
				if(map[i][j]=='S')
					goto start;
		
start:	
		for(f=0;f<4;f++)
			if(i+mm[f].x>-1&&i+mm[f].x<m&&j+mm[f].y>-1&&j+mm[f].y<n&&(map[i+mm[f].x][j+mm[f].y]=='.'||map[i+mm[f].x][j+mm[f].y]=='E')&&map[i+mm[f].x+mm[(f+3)%4].x][j+mm[f].y+mm[(f+3)%4].y]=='#')
			{	o=1+ldfs(i+mm[f].x,j+mm[f].y,f);		break;		}
			else if(i+mm[f].x>-1&&i+mm[f].x<m&&j+mm[f].y>-1&&j+mm[f].y<n&&map[i+mm[f].x][j+mm[f].y]=='.'&&(map[i+mm[f].x+mm[(f+3)%4].x][j+mm[f].y+mm[(f+3)%4].y]=='.'||map[i+mm[f].x+mm[(f+3)%4].x][j+mm[f].y+mm[(f+3)%4].y]=='E'))
			{	o=2+ldfs(i+mm[f].x+mm[(f+3)%4].x,j+mm[f].y+mm[(f+3)%4].y,(f+3)%4);		break;		}

		cout<<o<<' ';
		for(f=0;f<4;f++)
			if(i+mm[f].x>-1&&i+mm[f].x<m&&j+mm[f].y>-1&&j+mm[f].y<n&&(map[i+mm[f].x][j+mm[f].y]=='.'||map[i+mm[f].x][j+mm[f].y]=='E')&&map[i+mm[f].x+mm[(f+1)%4].x][j+mm[f].y+mm[(f+1)%4].y]=='#')
			{	o=1+rdfs(i+mm[f].x,j+mm[f].y,f);		break;		}
			else if(i+mm[f].x>-1&&i+mm[f].x<m&&j+mm[f].y>-1&&j+mm[f].y<n&&map[i+mm[f].x][j+mm[f].y]=='.'&&(map[i+mm[f].x+mm[(f+1)%4].x][j+mm[f].y+mm[(f+1)%4].y]=='.'||map[i+mm[f].x+mm[(f+1)%4].x][j+mm[f].y+mm[(f+1)%4].y]=='E'))
			{	o=2+rdfs(i+mm[f].x+mm[(f+1)%4].x,j+mm[f].y+mm[(f+1)%4].y,(f+1)%4);		break;		}

		cout<<o<<' ';
		lock[i][j]=1;	push(i);	push(j);	push(1);
		cout<<bfs()<<endl;
	}

	return 0;
}