poj 3083 Children of the Candy Corn 解題報告-- DFS BFS

題目大意：給出一個迷宮，#表示牆，S表示起點，E表示終點。求從S到E的優先靠左邊的牆走、優先靠右邊的牆走和S到E的最短路徑。

思路：考慮向左向右用DFS，最短路徑用BFS。注意幾點，一是儘量不用遞歸，開始用的遞歸，BFS會棧溢出，改用隊列實現後沒問題，而且其實邏輯上反而更統一，這裏DFS還好，遞歸沒有問題，大概是因爲有向左向右的規則，並不是一味地遍歷所有狀態；再一個就是BFS實在是狀態太多，必須限制，可以標記會走的點，不讓這些點重複地進入隊列。

當然，這個代碼明顯有很多可以修改的地方，在寫的過程中由於偷懶不影響結果的就沒改了。主要是練手而已，重在細心。

代碼：

#include<stdio.h>
#include <stdlib.h>
#include "queue"

using namespace std;

#define LEN 42//最外層不通

typedef struct {
	int x,y;
	int dx,dy;
	int layer;
} NODE;

char maze[LEN][LEN];
int w,h;
int ln,rn,sn;

void ini_maze(int *sx , int *sy)
{
	int i,j;

	for(i=0;i < LEN ; i++)
		for(j=0; j < LEN ; j++)
			maze[i][j]='#';

	for(i=1; i<=h; i++)
	{
		getchar();
		for(j=1; j<=w; j++)
		{
			scanf("%c",&maze[j][i]);
			if('S' == maze[j][i])
			{
				*sx = j;
				*sy = i;
			}
		}
	}

	ln = 1; rn =1; sn = 5001;
}

void find_sway(int x, int y, int dx, int dy , int layer)
{
	int lres = -1, rres = -1, mres = -1;
	NODE nod,td;
	queue<NODE> qlist;

	x += dx;
	y += dy;
	while(maze[x][y] != 'E')
	{
		if( maze[x+dy][y-dx] != '#')	//向左
		{
			nod.dx = dy;
			nod.dy = -dx;
			nod.x = x;
			nod.y = y;
			nod.layer = layer+1;
			qlist.push(nod);
			if(maze[x+dy][y-dx] != 'E')
				maze[x+dy][y-dx] = '#';
		}
		if( maze[x-dy][y+dx] != '#')	//向右
		{
			nod.dx = -dy;
			nod.dy = dx;
			nod.x = x;
			nod.y = y;
			nod.layer = layer+1;
			qlist.push(nod);
			if(maze[x-dy][y+dx] != 'E')
				maze[x-dy][y+dx] = '#';
		}
		if( maze[x+dx][y+dy] != '#')	//向前
		{
			nod.dx = dx;
			nod.dy = dy;
			nod.x = x;
			nod.y = y;
			nod.layer = layer+1;
			qlist.push(nod);
			if(maze[x+dx][y+dy] != 'E')
				maze[x+dx][y+dy] = '#';
		}
		td = qlist.front();
		x = td.x + td.dx;
		y = td.y + td.dy;
		dx = td.dx;
		dy = td.dy;
		layer = td.layer;
		qlist.pop();
	}
	
	sn = layer;
}

int find_lway(int x, int y, int dx, int dy , int layer)
{
	int lres = -1, rres = -1, mres = -1;

	ln++;
	x += dx;
	y += dy;
	if( maze[x][y] == 'E')
	{
		return layer;
	}
	
	if( maze[x+dy][y-dx] != '#')	//向左
	{
		lres = find_lway(x,y,dy,-dx,layer+1);
		if(lres == -1)
			ln++;
	}
	if(lres == -1)
	{
		if( maze[x+dx][y+dy] != '#')	//向前
		{
			mres = find_lway(x,y,dx,dy,layer+1);
			if(mres == -1)
				ln++;
		}
		if(mres == -1)
		{
			if( maze[x-dy][y+dx] != '#')	//向右
			{
				rres = find_lway(x,y,-dy,dx,layer+1);
				if(rres == -1)
					ln++;
			}
			if(rres == -1)
			{
				return -1;
			}
		}
	}
	return 0;
}

int find_rway(int x, int y, int dx, int dy , int layer)
{
	int lres = -1, rres = -1, mres = -1;

	rn++;
	x += dx;
	y += dy;
	if( maze[x][y] == 'E')
	{
		return layer;
	}
	
	if( maze[x-dy][y+dx] != '#')	//right
	{
		rres = find_rway(x,y,-dy,dx,layer+1);
		if(rres == -1)
			rn++;
	}
	if(rres == -1)
	{
		if( maze[x+dx][y+dy] != '#')	//向前
		{
			mres = find_rway(x,y,dx,dy,layer+1);
			if(mres == -1)
				rn++;
		}
		if(mres == -1)
		{
			if( maze[x+dy][y-dx] != '#')	//left
			{
				lres = find_rway(x,y,dy,-dx,layer+1);
				if(lres == -1)
					rn++;
			}
			if(lres == -1)
			{
				return -1;
			}
		}
	}
	return 0;
}

int main()
{
	int sx , sy,i,n;

//	freopen("in.txt","r",stdin);

	scanf("%d",&n);
	for(i=1; i<=n; i++)
	{
		scanf("%d%d",&w,&h);
		ini_maze(&sx,&sy);
		if(1 == sx)
		{
			find_lway(sx,sy,1,0,2);
			find_rway(sx,sy,1,0,2);
			find_sway(sx,sy,1,0,2);
		}
		if(1 == sy)
		{
			find_lway(sx,sy,0,1,2);
			find_rway(sx,sy,0,1,2);
			find_sway(sx,sy,0,1,2);
		}
		if(w == sx)
		{
			find_lway(sx,sy,-1,0,2);
			find_rway(sx,sy,-1,0,2);
			find_sway(sx,sy,-1,0,2);
		}
		if(h == sy)
		{
			find_lway(sx,sy,0,-1,2);
			find_rway(sx,sy,0,-1,2);
			find_sway(sx,sy,0,-1,2);
		}
		printf("%d %d %d\n",ln,rn,sn);
	}
	system("pause");

	return 0;
}