POJ 3984 迷宫问题 --- BFS

 

 

/* POJ 3984 迷宫问题 --- BFS */
#include <cstdio>

int mapn[5][5];
int x[100], y[100], pre[100];
int front = 0;//队首
int rear = 1;//队尾指向首元素的后一位
int dx[4] = { 1, -1, 0, 0 }; //4个方向
int dy[4] = { 0, 0, 1, -1 };

void printmap(int i){
    //注意pre[0]的值为-1，所以(0,0)没有输出
    if (pre[i] != -1){
        printmap(pre[i]);
        printf("(%d, %d)\n", x[i], y[i]);
    }
}

void bfs(int x0, int y0){
    x[front] = x0;
    y[front] = y0;
    pre[front] = -1;  //首元素(即左上角元素)入队

    while (front < rear){
        //队不为空时
        for (int i = 0; i < 4; ++i){
            //往四个方向广搜,判断是否可以访问
            int a = x[front] + dx[i];
            int b = y[front] + dy[i];
            if (a < 0 || a > 4 || b < 0 || b > 4 || mapn[a][b] == 1)
                continue; //无法访问的点
            else{
                mapn[a][b] = 1; //第一次访问就是最短的，访问后标记访问过(直接改为1可以省去一个visit数组)
                //满足的元素入队
                x[rear] = a;
                y[rear] = b;
                pre[rear] = front; //记录路径(最短路常用)
                ++rear;
                if (a == 4 && b == 4)
                    printmap(front); //(4,4)所在的点的下标此时并未入队，故(4，4)也未输出
            }
        }//for(i);
        ++front; //一访问过的元素出队
    }
}

int main()
{
    for (int i = 0; i < 5; ++i){
        for (int j = 0; j < 5; ++j){
            scanf("%d", mapn[i] + j);
        }//for(j)
    }//for(i)
    printf("(0, 0)\n");
    bfs(0, 0);
    printf("(4, 4)\n");

    return 0;
}

 

用结构体似乎更好理解一点：

#include <cstdio>

const int maxn = 5;
int mapp[maxn][maxn];
int dx[4] = { -1, 1, 0, 0 };
int dy[4] = { 0, 0, -1, 1 };

struct point{
    int x, y, pre;
    point(int a, int b, int c) :x(a), y(b), pre(c){}
    point(){}
}q[105];
int head, rear;

void printmap(int i){
    if (i == -1)
        return;
    else{
        printmap(q[i].pre);
        printf("(%d, %d)\n", q[i].x, q[i].y);
    }

}

void bfs(int x0, int y0){
    head = 0, rear = 1;
    //首元素入队
    q[head].x = x0;
    q[head].y = y0;
    q[head].pre = -1;

    while (head < rear){
        point tmp = q[head];
        for (int i = 0; i < 4; ++i){
            int x = tmp.x + dx[i];
            int y = tmp.y + dy[i];
            if (x < 0 || x >= 5 || y < 0 || y >= 5 || mapp[x][y] == 1)
                continue;
            else{
                mapp[x][y] = 1;
                q[rear] = point(x, y, head);
                ++rear;
            }
            if (x == 4 && y == 4)
                printmap(rear-1);

        }
        ++head;
    }
}

int main()
{
    for (int i = 0; i < 5; ++i){
        for (int j = 0; j < 5; ++j){
            scanf("%d", mapp[i] + j);
        }
    }
    bfs(0, 0);

    return 0;
}