[Programming thinking and practice Week2 job A Maze]

Meaning of the questions:

There is a map, the map in the upper left corner is the entrance, the lower right corner for export. Map, 1 can not walk, 0 can go. To ensure that import, export zero. Write a program to find the entrance to the exit of the shortest route.

Input:

Input is a 5 × 5 two-dimensional array, just the two numbers 0,1, representation positions in FIG.
The Input the Sample:
0. 1 0 0 0
0 0. 1. 1 0
0 0. 1. 1 0
0 0 0. 1 0
0 0. 1. 1 0

Output:

A plurality of output lines, represents the shortest path from upper left to lower right corner coordinates sequentially passes. Ensure data has a unique solution.
The Sample the Output:
(0, 0)
(. 1, 0)
(2, 0)
(. 3, 0)
(. 3,. 1)
(. 3, 2)
(2, 2)
(. 1, 2)
(0, 2)
(0 , 3)
(0, 4)
(1, 4)
(2, 4)
(3, 4)
(4, 4)

Ideas:

The map is initialized to an array of 5 * 5, with a map showing the obstacles, with 0 map to reach the place but has not yet arrived. From the start breadth-first search (the BFS), has been used since the number 0, so the use of digital> 2 marking the path length (ie start tag 2, the starting point can be reached around a point labeled 3, this and so on), the search stops encountered exit. Finally, according to exit from the record marked paths (each looking to record a small path length than the current point of a point) until you return to the starting point, find the shortest path.

Code:

#include <iostream>
#include<stdio.h>
#include<queue>
#include<vector>
using namespace std;
const int maxn=5;
int vis[maxn][maxn];
int dx[]={1,-1,0,0},
	dy[]={0,0,1,-1};
int sx=0,sy=0,tx=4,ty=4;
struct Point{
	int x;
	int y;
	Point(int thex,int they)
	{
		x=thex,y=they;
	}
	
};
queue<Point> q;
void bfs()
{
	q.push(Point(sx,sy));
	vis[sx][sy]=2;
	while(!q.empty())
	{
		Point now=q.front();
		q.pop();
		for(int i=0;i<4;i++)
		{
			int x=now.x+dx[i];
			int y=now.y+dy[i];
			if(x>=0&&x<maxn&&y>=0&&y<maxn&&vis[x][y]==0)
			{
				vis[x][y]=vis[now.x][now.y]+1;
				q.push(Point(x,y));
				if(x==tx&&y==ty)
					return;
			}
		}
	}
}
void print()
{
	vector<Point> v;
	Point now(tx,ty);
	v.push_back(Point(tx,ty));
	while(true)
	{
		if(now.x==sx&&now.y==sy)
		{
			//printf("(%d,%d)\n",sx,sy);
			break;
		}
		for(int i=0;i<4;i++)
		{
			int x=now.x+dx[i],y=now.y+dy[i];
			if(x>=0&&x<maxn&&y>=0&&y<maxn&&vis[x][y]==vis[now.x][now.y]-1)
			{
				v.push_back(Point(x,y));
				now.x=x,now.y=y;
				break;
			}
		}
	}
	vector<Point>::reverse_iterator it=v.rbegin();
	printf("(%d, %d)",it->x,it->y);
	it++;
	for(;it!=v.rend();it++)
		printf("\n(%d, %d)",it->x,it->y);
}
int main(int argc, char** argv) {
	for(int i=0;i<maxn;i++)
		for(int j=0;j<maxn;j++)
			cin>>vis[i][j];
	bfs();
	print();
	return 0;
}