Basic wall maze

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Special Judge

Problem Description

In this problem you have to solve a very simple maze consisting of:

1.a 6 by 6 grid of unit squares

2.3 walls of length between 1 and 6 which are placed either horizontally or vertically to separate squares

3.one start and one end marker

A maze may look like this:

You have to find a shortest path between the square with the start marker and the square with the end marker. Only moves between adjacent grid squares are allowed; adjacent means that the grid squares share an edge and are not separated by a wall. It is not
allowed to leave the grid.

Input

The input consists of several test cases. Each test case consists of five lines: The first line contains the column and row number of the square with the start marker, the second line the column and row number of the square with the end marker. The third, fourth
and fifth lines specify the locations of the three walls. The location of a wall is specified by either the position of its left end point followed by the position of its right end point (in case of a horizontal wall) or the position of its upper end point
followed by the position of its lower end point (in case of a vertical wall). The position of a wall end point is given as the distance from the left side of the grid followed by the distance from the upper side of the grid.

You may assume that the three walls don‘t intersect with each other, although they may touch at some grid corner, and that the wall endpoints are on the grid. Moreover, there will always be a valid path from the start marker to the end marker. Note that the
sample input specifies the maze from the picture above.

The last test case is followed by a line containing two zeros.

Output

For each test case print a description of a shortest path from the start marker to the end marker. The description should specify the direction of every move (‘N‘ for up, ‘E‘ for right, ‘S‘ for down and ‘W‘ for left).

There can be more than one shortest path, in this case you can print any of them.

Sample Input

1 6
2 6
0 0 1 0
1 5 1 6
1 5 3 5
0 0

Sample Output

NEEESWW

题意：给定一张地图，并且给定起点和终点，求起点到终点的最短距离，地图上有墙，与以往的题目不同的是，以往的题目障碍物都是在格子上，但是本题的障碍物墙是在格子与格子的边界线上，所以在输入的时候就要进行预处理下，将墙的位置转化为相邻格子的东西南北方向墙的状态，所以使用了一个3为数组来记录地图的信息map[x][y][0]-map[x][y][3] 分别表示坐标为x，y的格子的四个方向墙的情况，0为没墙，1为有墙，然后一个dfs找到最短路，以及每个点的前驱节点，最后打印路径。代码中的注释很详细。题目本身很简单，就是代码写起来有点麻烦。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <stack>

using namespace std;

const int MAX = 9,limit = 6,INF = 1000;
const int dirx[4]={0,-1,0,1},diry[4]={1,0,-1,0};

//map[x][y][0]-map[x][y][3] 分别表示坐标为x，y的格子的四个方向墙的情况，0为没墙，1为有墙
int map[MAX][MAX][4];
//pre[x][y][0]用来记录x，y的前驱格子的x坐标，pre[x][y][1]用来记录x，y的前驱格子的y坐标
int dist[MAX][MAX],pre[MAX][MAX][2];
int sx,sy,ex,ey,pax,pay,pbx,pby;
stack<char> st;

void init(){
	int i,j;

	for(i=0;i<MAX;++i){
		for(j=0;j<MAX;++j){
			dist[i][j] = INF;
			map[i][j][0] = map[i][j][1] = map[i][j][2] = map[i][j][3] = 0;
		}
	}
}

void dfs(int x,int y,int cnt){

	int i,tx,ty;

	for(i=0;i<4;++i){
		if(map[x][y][i]==1)continue;
		tx = x+dirx[i];
		ty = y+diry[i];
		if(tx<1 || ty<1 || tx>limit || ty>limit)continue;
		if(cnt+1>dist[tx][ty])continue;
		//更短就要更新，并且记录前驱
		dist[tx][ty] = cnt;
		pre[tx][ty][0] = x;
		pre[tx][ty][1] = y;
		dfs(tx,ty,cnt+1);
	}
}

void Path(){
	int px,py,x,y;

	x = ex,y = ey;
	px = pre[x][y][0];
	py = pre[x][y][1];

	while(1){
		//判断方向
		if(x==px){//x坐标相同看y坐标的情况
			if(py<y)st.push(‘E‘);
			else st.push(‘W‘);
		}else{//y坐标相同看x坐标的情况
			if(px<x)st.push(‘S‘);
			else st.push(‘N‘);
		}
		if(px==sx && py==sy)break;
		x = px;
		y = py;
		px = pre[x][y][0];
		py = pre[x][y][1];
	}

	while(!st.empty()){
		printf("%c",st.top());
		st.pop();
	}
	printf("\n");
}

int main(){
    //freopen("in.txt","r",stdin);
    //(author : CSDN iaccepted)
	int i,j;
	while(scanf("%d %d",&sy,&sx)){
		if(sx==0 && sy==0)break;
		scanf("%d %d",&ey,&ex);

		init();
		for(i=0;i<3;++i){
			scanf("%d %d %d %d",&pay,&pax,&pby,&pbx);
			if(pax==pbx){
				for(j=pay+1;j<=pby;++j){
					map[pax][j][3] = 1;
					map[pax+1][j][1] = 1;
				}
			}else{
				for(j=pax+1;j<=pbx;++j){
					map[j][pby][0] = 1;
					map[j][pby+1][2] = 1;
				}
			}
		}

		dfs(sx,sy,0);

		Path();
	}

    return 0;
}

HDU 1484 Basic wall maze (dfs + 记忆化)