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 00 1 0 1 00 0 0 0 00 1 1 1 00 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;}