USACO 2.4 Overfencing(BFS)

简单的BFS，可是看错题了，从原来的基础上改的，改的很乱，边界什么的。错了好几次。

  1 /*
  2   ID: cuizhe
  3   LANG: C++
  4   TASK: maze1
  5 */
  6 #include <iostream>
  7 #include <cstdio>
  8 #include <cstring>
  9 #include <cmath>
 10 #include <algorithm>
 11 using namespace std;
 12 char p[501][501];
 13 int o[501][501];
 14 int quer[100000],quec[100000];
 15 int x[4] = {0,0,-1,1};
 16 int y[4] = {1,-1,0,0};
 17 int main()
 18 {
 19     int w,h,j,i,k,str,end,num;
 20     FILE *fp;
 21     fp = freopen("maze1.in","r",stdin);
 22     freopen("maze1.out","w",stdout);
 23     scanf("%d%d%*c",&w,&h);
 24     for(i = 0; i <= 2*h; i ++)
 25     {
 26         fgets(p[i],2*w+3,fp);
 27     }
 28     str = end = 1;
 29     for(i = 0; i <= 2*h; i ++)
 30     {
 31         if(p[i][0] == ' ')
 32         {
 33             quer[end] = i;
 34             quec[end] = 0;
 35             o[i+1][1] = 1;
 36             end++;
 37         }
 38         if(p[i][2*w] == ' ')
 39         {
 40             quer[end] = i;
 41             quec[end] = 2*w;
 42             o[i+1][2*w+1] = 1;
 43             end++;
 44         }
 45     }
 46     for(i = 0; i <= 2*w; i ++)
 47     {
 48         if(p[0][i] == ' ')
 49         {
 50             quer[end] = 0;
 51             quec[end] = i;
 52             o[1][i+1] = 1;
 53             end++;
 54         }
 55         if(p[2*h][i] == ' ')
 56         {
 57             quer[end] = 2*h;
 58             quec[end] = i;
 59             o[2*h+1][i+1] = 1;
 60             end++;
 61         }
 62     }
 63     end --;
 64     num = 0;
 65     while(str <= end)
 66     {
 67         j = 1;
 68         for(i = str; i <= end; i ++)
 69         {
 70             for(k = 0; k <= 3; k ++)
 71             {
 72                 int rr,cc;
 73                 rr = quer[i]+x[k];
 74                 cc = quec[i]+y[k];
 75                 if(rr<=2*h&&rr>=0&&cc<=2*w&&cc>=0)
 76                 {
 77                     if(num != 0)
 78                     {
 79                         if(rr+x[k]<=2*h&&rr+x[k]>=0&&cc+y[k]<=2*w&&cc+y[k]>=0)
 80                         {
 81                             if(p[rr][cc] == ' '&&!o[rr+x[k]][cc+y[k]])
 82                             {
 83                                 o[rr+x[k]][cc+y[k]] = num+1;
 84                                 quer[end+j] = rr+x[k];
 85                                 quec[end+j] = cc+y[k];
 86                                 j ++;
 87                             }
 88                         }
 89                     }
 90                     else
 91                     {
 92                         if(p[rr][cc] == ' '&&!o[rr][cc])
 93                         {
 94                             o[rr][cc] = num+1;
 95                             quer[end+j] = rr;
 96                             quec[end+j] = cc;
 97                             j ++;
 98                         }
 99                     }
100                 }
101             }
102         }
103         str = end+1;
104         end = end+j-1;
105         num ++;
106     }
107     printf("%d\n",num-1);
108     return 0;
109 }
posted @ 2012-11-14 19:55  Naix_x  阅读(198)  评论(0)  编辑  收藏  举报