POJ9384 迷宫问题（基础BFS)

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本文出自：http://blog.csdn.net/svitter

题目：让你从(0, 0)走到（4,4)，并且输出路径。输入数据：二位数组的迷宫；输出数据：路径；

题解：简单的BFS

注意：

1.去重；

2.墙不能走；

3.记录前一个节点

代码：

#include <iostream>#include <stdio.h>#include <string.h>using namespace std;int maze[6][6];bool visit[5][5];struct node{    node *pre;    int i;    int j;    node(){}    node(int a, int b, node*c):i(a), j(b), pre(c){}};//1.去重//2.是否墙壁//inline bool judge(int i, int j, int temp){    if(temp < 0 || temp > 4)        return false;    //i, j 可以通行，并且没有访问过    if(maze[i][j] != 1 && !visit[i][j])        return true;    return false;}void output(){}int main(){    int i, j;    for(i = 0; i < 5; i++)        for(j = 0; j < 5; j++)            scanf("%d", &maze[i][j]);    node queue[1000];    memset(visit, 0, sizeof(visit));    int front , rear;    front = rear = -1;    node cur;    int temp;    queue[++rear] = node(4, 4, NULL);    while(front != rear)    {        cur = queue[++front];        i = cur.i;        j = cur.j;        if(i == 0 && j == 0)            break;        temp = cur.i+1;        if(judge(temp, j, temp))        {            queue[++rear] = node(temp, j, &queue[front]);            visit[temp][j] = 1;        }        temp = cur.j+1;        if(judge(i, temp, temp))        {            queue[++rear] = node(i, temp, &queue[front]);            visit[i][temp] = 1;        }        temp = cur.i-1;        if(judge(temp, j, temp))        {            queue[++rear] = node(temp, j, &queue[front]);            visit[temp][j] = 1;        }        temp = cur .j-1;        if(judge(i, temp, temp))        {            queue[++rear] = node(i, temp, &queue[front]);            visit[i][temp] = 1;        }    }    node *tmp;    tmp = &queue[front];    while(tmp != NULL)    {        printf("(%d, %d)\n", tmp->i, tmp->j);        tmp = tmp->pre;    }    return 0;}

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