• COJ 1080 A simple maze


    地图很小,50×50,所以普通的BFS就行了;

    1WA:地图用%s读取时,读入的是字符0,在判断是否有路径时和数值0比较。

     1 # include <stdio.h>
     2 # include <string.h>
     3 
     4 typedef struct {short x, y;}queue;
     5 
     6 const short dir[4][2] = {{-1,0}, {1,0}, {0,-1}, {0,1}};
     7 
     8 int r, c, sr, sc, er, ec;
     9 char m[55][55], v[55][55], d[55][55];
    10 queue Q[55*55];
    11 
    12 void read(void)
    13 {
    14     int i;
    15     
    16     scanf("%d%d%d%d%d%d", &r, &c, &sr, &sc, &er, &ec);
    17     for (i = 1; i <= r; ++i)
    18         scanf("%s", m[i]+1);
    19         
    20     /* 设置边界 */
    21     for (i = 0; i <= c; ++i)
    22     {
    23         m[0][i] = 1;
    24         m[r+1][i+1] = 1;
    25     }
    26     for (i = 0; i <= r; ++i)
    27     {
    28         m[i][c+1] = 1;
    29         m[i+1][0] = 1;
    30     }
    31 }
    32 
    33 void solve(void)
    34 {
    35     int front, rear, ans, i;
    36     queue cur, nst;
    37     
    38     ans = -1;
    39     
    40     memset(Q, 0, sizeof(Q));
    41     memset(v, 0, sizeof(v));
    42     memset(d, 0, sizeof(d));
    43     
    44     Q[1].x = sr, Q[1].y = sc;
    45     v[sr][sc] = 1;
    46     front = 1, rear = 2;
    47     while (front < rear)
    48     {
    49         cur = Q[front++];
    50         if (cur.x == er && cur.y == ec) {ans = d[cur.x][cur.y]; break;}
    51         for (i = 0; i < 4; ++i)
    52         {
    53             nst.x = cur.x + dir[i][0];
    54             nst.y = cur.y + dir[i][1];
    55             if (nst.x>=1 && nst.x<=r && nst.y>=1 && nst.y<=c)
    56             {
    57                 if (m[nst.x][nst.y]=='0' && !v[nst.x][nst.y])
    58                 {
    59                     Q[rear++] = nst;
    60                     v[nst.x][nst.y] = 1;
    61                     d[nst.x][nst.y] = d[cur.x][cur.y] + 1;
    62                 }
    63             }
    64         }
    65     }
    66     
    67     if (ans == -1) puts("No Path.");
    68     else printf("%d\n", ans);
    69 
    70 }
    71 
    72 int main()
    73 {
    74     int T;
    75         
    76     scanf("%d", &T);
    77     while (T--)
    78     {
    79         read();
    80         solve();
    81     }
    82     
    83     return 0;
    84 }

    //

  • 相关阅读:
    Hello CnBlog
    boxshadow
    箭头函数
    Web存储机制
    闭包
    HTTP Cookie
    javascript执行环境和作用域链
    BFC
    iOS开发之创建颜色渐变视图View
    iOS之中国银联移动支付控件升级的问题
  • 原文地址:https://www.cnblogs.com/JMDWQ/p/2515073.html
Copyright © 2020-2023  润新知