POJ 2935 Basic Wall Maze

1 算法

简单的BFS，但WA了好久。

队列q保存BFS过程中的节点的坐标（x,y)，父节点的地址father， 和父节点到达当前节点所走的方向c， 输出路径时根据目标节点的father索引回源节点。

2 代码

2.1 WA的代码

#Problem: 2935#Memory: N/ATime: N/A#Language: G++Result: Wrong Answer#include <cstdio>#include <iostream>#include <cstring>using namespace std;struct way{        struct way *father;        int x, y;        char c;};int sx, sy;int tx, ty;struct way q[100];int qf, ql;int offsetx[4] = {0, 0, 1, -1};int offsety[4] = {-1, 1, 0, 0};bool wall[8][8][8][8];char direction[4] = {'N', 'S', 'E', 'W'};bool isv[7][7];void read_data(){        int x1, y1, x2, y2;        cin >> tx >> ty;        memset(wall, 0, sizeof(wall));        for (int i = 0; i < 3; ++i)        {                cin >> x1 >> y1 >> x2 >> y2;                if (x1 == x2)                {                        if (y1 > y2) swap(y1, y2);                        for (int k = y1 + 1; k <= y2; ++k)                        {                                wall[x1][k][x1 + 1][k] = true;                                wall[x1 + 1][k][x1][k] = true;                        }                }                if (y1 == y2)                {                        if (x1 > x2) swap(x1, x2);                        for (int k = x1 + 1; k <= x2; ++k)                        {                                wall[k][y1][k][y1 + 1] = true;                                wall[k][y1 + 1][k][y1] = true;                        }                }        }}bool isinside(int x, int y){        return (1 <= x && x < 7 && 1 <= y && y < 7);}void print(struct way *p){        if (p == NULL) return ;        print(p->father);        printf("%c", p->c);}void bfs(){        int fx, fy;        int nx, ny;        qf = ql = 0;        q[ql].father = NULL;        q[ql].x = sx;        q[ql].y = sy;        memset(isv, 0, sizeof(isv));        isv[sx][sy] = true;        ql++;        while (qf != ql)        {                fx = q[qf].x;                fy = q[qf].y;                for (int i = 0; i < 4; ++i)                {                        nx = fx + offsetx[i];                        ny = fy + offsety[i];                        if (isinside(nx, ny) && !isv[nx][ny] &&                                         !wall[nx][ny][fx][fy])                        {                               q[ql].x = nx;                               q[ql].y = ny;                               q[ql].father = &(q[qf]);                               q[ql].c = direction[i];                               if (nx == tx && ny == ty)                               {                                       print(&(q[ql]));                                       return ;                               }                               ql++;                               isv[nx][ny] = true;                        }                }                qf++;        }}int main(){        cin >> sx >> sy;        while (sx != 0 && sy != 0)        {                read_data();                bfs();                printf("\n");                cin >> sx >> sy;        }        return 0;}

2.2 修改

62行

        if (p == NULL) return ;

改为

        if (p->father == NULL) return ;