当前位置：网站首页>poj2935 Basic Wall Maze （2016xynu暑期集训检测 -----D题）

poj2935 Basic Wall Maze （2016xynu暑期集训检测 -----D题）

2022-08-05 10:26:00 【51CTO】

Description

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

a 6 by 6 grid of unit squares
3 walls of length between 1 and 6 which are placed either horizontally or vertically to separate squares
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

如果去掉障碍那么就是一道dijkstra的最短路径题

既然有了障碍那么就处理障碍

把障碍相隔的两个点设置为不能通过

剩下就是bfs了

       
       #include <stdio.h>
       
       #include <string.h>
       
       #include <queue>
       
       #include <algorithm>
       
       using 
       
       namespace 
       
       std;
       
       int 
       
       st_x,
       
       st_y;
       
       int 
       
       ed_x,
       
       ed_y;
       
       //wall[i][j][k][l] 表示(i,j)与(k,l)之间有障碍 
       
       bool 
       
       wall[
       
       10][
       
       10][
       
       10][
       
       10];
       
       int 
       
       dir[
       
       4][
       
       2]
       
       ={
       
       1,
       
       0,
       
       -
       
       1,
       
       0,
       
       0,
       
       1,
       
       0,
       
       -
       
       1};
       
       char 
       
       Dir[
       
       5]
       
       ={
       
       "EWSN"};
       
       bool 
       
       vis[
       
       10][
       
       10];
       
       bool 
       
       judge(
       
       int 
       
       x,
       
       int 
       
       y)
       
{
       
       if(
       
       x
       
       <=
       
       0
       
       ||
       
       y
       
       <=
       
       0
       
       ||
       
       x
       
       >
       
       6
       
       ||
       
       y
       
       >
       
       6
       
       ||
       
       vis[
       
       x][
       
       y])
       
       return 
       
       false;
       
       return 
       
       true;
       
}
       
       struct 
       
       node
       
{
       
       int 
       
       step;
       
       int 
       
       index;
       
       int 
       
       x,
       
       y;
       
       char 
       
       path[
       
       100];
       
       friend 
       
       bool 
       
       operator
       
       <(
       
       node 
       
       x1,
       
       node 
       
       y1)
       
  {
       
       return 
       
       x1.
       
       step
       
       >
       
       y1.
       
       step;
       
  }
       
};
       
       void 
       
       bfs()
       
{
       
       memset(
       
       vis,
       
       false,
       
       sizeof(
       
       vis));
       
       priority_queue
       
       <
       
       node
       
       >
       
       s;
       
       node 
       
       temp1,
       
       temp2;
       
       temp1.
       
       x
       
       =
       
       st_x;
       
       temp1.
       
       y
       
       =
       
       st_y;
       
       temp1.
       
       step
       
       =
       
       0;
       
       temp1.
       
       index
       
       =
       
       0;
       
       s.
       
       push(
       
       temp1);
       
       while(
       
       !
       
       s.
       
       empty())
       
  {
       
       temp1
       
       =
       
       temp2
       
       =
       
       s.
       
       top();
       
       s.
       
       pop();
       
       if(
       
       temp1.
       
       x
       
       ==
       
       ed_x
       
       &&
       
       temp1.
       
       y
       
       ==
       
       ed_y)
       
       break;
       
       if(
       
       vis[
       
       temp1.
       
       x][
       
       temp1.
       
       y])
       
       continue;
       
       vis[
       
       temp1.
       
       x][
       
       temp1.
       
       y]
       
       =
       
       true;
       
       for(
       
       int 
       
       i
       
       =
       
       0;
       
       i
       
       <
       
       4;
       
       i
       
       ++)
       
    {
       
       int 
       
       xx
       
       =
       
       temp1.
       
       x
       
       +
       
       dir[
       
       i][
       
       0];
       
       int 
       
       yy
       
       =
       
       temp1.
       
       y
       
       +
       
       dir[
       
       i][
       
       1];
       
       if(
       
       judge(
       
       xx,
       
       yy)
       
       &&!
       
       wall[
       
       temp1.
       
       x][
       
       temp1.
       
       y][
       
       xx][
       
       yy])
       
      {
       
       temp1.
       
       x
       
       =
       
       xx;
       
       temp1.
       
       y
       
       =
       
       yy;
       
       temp1.
       
       path[
       
       temp1.
       
       index
       
       ++]
       
       =
       
       Dir[
       
       i];
       
       temp1.
       
       path[
       
       temp1.
       
       index]
       
       =
       
       '\0';
       
       temp1.
       
       step
       
       ++;
       
       s.
       
       push(
       
       temp1);
       
      }
       
       temp1
       
       =
       
       temp2;
       
    }
       
  }
       
       while(
       
       !
       
       s.
       
       empty())
       
       s.
       
       pop();
       
       printf(
       
       "%s\n",
       
       temp1.
       
       path);
       
}
       
       int 
       
       main()
       
{
       
       while(
       
       ~scanf(
       
       "%d %d",
       
       &
       
       st_x,
       
       &
       
       st_y))
       
  {
       
       if(
       
       !
       
       st_x
       
       &&!
       
       st_y) 
       
       break;
       
       scanf(
       
       "%d %d",
       
       &
       
       ed_x,
       
       &
       
       ed_y);
       
       memset(
       
       wall,
       
       false,
       
       sizeof(
       
       wall));
       
       for(
       
       int 
       
       i
       
       =
       
       0;
       
       i
       
       <
       
       3;
       
       i
       
       ++)
       
    {
       
       int 
       
       a,
       
       b,
       
       c,
       
       d;
       
       scanf(
       
       "%d %d %d %d",
       
       &
       
       a,
       
       &
       
       b,
       
       &
       
       c,
       
       &
       
       d);
       
       //处理障碍 
       
       if(
       
       a
       
       ==
       
       c)
       
      {
       
       int 
       
       l
       
       =
       
       min(
       
       b,
       
       d);
       
       int 
       
       r
       
       =
       
       max(
       
       b,
       
       d);
       
       for(
       
       int 
       
       j
       
       =
       
       l
       
       +
       
       1;
       
       j
       
       <=
       
       r;
       
       j
       
       ++)
       
        {
       
       wall[
       
       a][
       
       j][
       
       a
       
       +
       
       1][
       
       j]
       
       =
       
       true;
       
       wall[
       
       a
       
       +
       
       1][
       
       j][
       
       a][
       
       j]
       
       =
       
       true;
       
        }
       
      }
       
       //处理障碍 
       
       if(
       
       b
       
       ==
       
       d)
       
      {
       
       int 
       
       l
       
       =
       
       min(
       
       a,
       
       c);
       
       int 
       
       r
       
       =
       
       max(
       
       a,
       
       c);
       
       for(
       
       int 
       
       j
       
       =
       
       l;
       
       j
       
       <
       
       r;
       
       j
       
       ++)
       
        {
       
       wall[
       
       j
       
       +
       
       1][
       
       b][
       
       j
       
       +
       
       1][
       
       b
       
       +
       
       1]
       
       =
       
       true;
       
       wall[
       
       j
       
       +
       
       1][
       
       b
       
       +
       
       1][
       
       j
       
       +
       
       1][
       
       b]
       
       =
       
       true;
       
        }
       
      }
       
    }
       
       bfs();
       
  }
       
       return 
       
       0;
       
}
      
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版权声明
本文为[51CTO]所创，转载请带上原文链接，感谢
https://blog.51cto.com/u_15742657/5546711

当前位置：网站首页>poj2935 Basic Wall Maze （2016xynu暑期集训检测 -----D题）

poj2935 Basic Wall Maze （2016xynu暑期集训检测 -----D题）

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