poj3984迷宫问题 广搜+最短路径

绝对经典！

定义一个二维数组： 

int maze[5][5] = {
0, 1, 0, 0, 0,
0, 1, 0, 1, 0,
0, 0, 0, 0, 0,
0, 1, 1, 1, 0,
0, 0, 0, 1, 0,
};

它表示一个迷宫，其中的1表示墙壁，0表示可以走的路，只能横着走或竖着走，不能斜着走，要求编程序找出从左上角到右下角的最短路线。

Input

一个5 × 5的二维数组，表示一个迷宫。数据保证有唯一解。

Output

左上角到右下角的最短路径，格式如样例所示。

Sample Input

0 1 0 0 0
0 1 0 1 0
0 0 0 0 0
0 1 1 1 0
0 0 0 1 0

Sample Output

(0, 0)
(1, 0)
(2, 0)
(2, 1)
(2, 2)
(2, 3)
(2, 4)
(3, 4)
(4, 4)

方法一：

<pre name="code" class="cpp">#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <queue>
using namespace std;
struct Node
{
	int x,y;
}node[30];
int father[30],map[5][5],vis[5][5];
int xx[4]={1,-1,0,0};
int yy[4]={0,0,1,-1};
queue<Node>q;
void printf_1()
{
	int a[30];
	//printf("(0, 0)\n");
	int u=24;
	int j=0;
	while(1)
	{
		a[j++]=father[u];
		u=father[u];
		if(u==0)
			break;
	}
	//printf("(0, 0)\n");
	for(int i=j-1;i>=0;i--)
		printf("(%d, %d)\n",a[i]/5,a[i]%5);
	printf("(4, 4)\n");
}
void bfs(int x,int y)
{
	memset(vis,0,sizeof(vis));
	memset(father,-1,sizeof(father));
	vis[0][0]=1;
	Node now,next;
	now.x=0,now.y=0;
	//father[x*5+y]=0;
	q.push(now);
	while(!q.empty())
	{
		now=q.front();
		q.pop();
		for(int i=0;i<4;i++)
		{
			int dx=now.x+xx[i];
			int dy=now.y+yy[i];
			if(dx>=0&&dx<=4&&dy>=0&&dy<=4&&vis[dx][dy]==0&&map[dx][dy]==0)
			{
				vis[dx][dy]=1;
				father[dx*5+dy]=now.x*5+now.y;
				if(dx==4&&dy==4)
				{
					printf_1();
					return ;
				}
				next.x=dx;
				next.y=dy;
				q.push(next);
			}
		}
	}
}
int main ()
{
	for(int i=0;i<=4;i++)
		for(int j=0;j<=4;j++)
			scanf("%d",&map[i][j]);
	bfs(0,0);
	return 0;
}

方法二：模拟队列

<pre name="code" class="cpp">#include <iostream>
#include <string.h>
#include <fstream>
using namespace std;

struct node {
	int x;
	int y;
	int pre;
}edge[100];

int front = 0, rear = 1;     //队头，队尾
int map[5][5];
int vis[5][5];
int d[4][2] = {{1,0},{0,-1},{-1,0},{0,1}};

void PrePath(int i) {          //倒向追踪法
	if(edge[i].pre != -1) {
		PrePath(edge[i].pre);
		cout<<"("<<edge[i].x<<", "<<edge[i].y<<")"<<endl;
	}
}

void bfs(int x1, int y1, int x2, int y2) {
	edge[front].x = x1;
	edge[front].y = y1;
	edge[front].pre = -1;
	while(front < rear) {               //队列为空时终止
		for(int u = 0; u < 4; u++) {
			int x = edge[front].x + d[u][0];
			int y = edge[front].y + d[u][1];
			if((x < 0) || (x >= 5) || (y < 0) || (y >= 5) || (map[x][y] == 1) || (vis[x][y] == 1)) {
				continue;
			} else {
				vis[x][y] = 1;
				map[x][y] = 1;
				edge[rear].x = x;    //入队
				edge[rear].y = y;
				edge[rear].pre = front;
				rear++;
			}
			if(x == x2 && y == y2) {
				PrePath(front);
			}
		}
		front++;     //出队
	}
}

int main() {
	freopen("C:\\Users\\HZH\\Desktop\\input.txt","r",stdin);
	memset(vis, 0, sizeof(vis));
	for(int i = 0; i < 5; i++) {
		for(int j = 0; j < 5; j++) {
			cin>>map[i][j];
		}
	}
	int x1,x2,y1,y2;
	cin>>x1>>y1>>x2>>y2;
	fclose(stdin);
	cout<<"("<<x1<<", "<<y1<<")"<<endl;
	bfs(x1, y1, x2, y2);
	cout<<"("<<x2<<", "<<y2<<")"<<endl;
}