USACO 3.2 Magic Squares(记录路径的BFS）

复杂，这种BFS，就是模拟的节奏啊，写了好久。用8进制记录每一个状态，然后顺序一定要想清楚，然后变换的时候注意一下。很多细节问题，导致我调试了很久。

  1 /*
  2      ID: cuizhe
  3      LANG: C++
  4      TASK: msquare
  5 */
  6 #include <cstdio>
  7 #include <cstring>
  8 #include <string>
  9 #include <cmath>
 10 #include <queue>
 11 #include <map>
 12 using namespace std;
 13 int o[10],p[10],que[100000],pre[100000],st[100000];
 14 map<int,int>mp;
 15 void judge(int x)
 16 {
 17     int num = 1;
 18     memset(p,0,sizeof(p));
 19     while(x)
 20     {
 21         p[num++] = x%8;
 22         x = x/8;
 23     }
 24 }
 25 int main()
 26 {
 27     int i,j,k,n,aim,str,end,sum,num = 0,tail;
 28     int t[10];
 29     char ch[1001];
 30     freopen("msquare.in","r",stdin);
 31     freopen("msquare.out","w",stdout);
 32     aim = 0;
 33     o[1] = 1;
 34     for(i = 2; i <= 8; i ++)
 35     {
 36         o[i] = o[i-1]*8;
 37     }
 38     for(i = 1; i <= 8; i ++)
 39     {
 40         scanf("%d",&n);
 41         aim += (n-1)*o[i];
 42     }
 43     que[1] = 16434824;
 44     mp[que[1]] = 1;
 45     str = end = 1;
 46     while(str <= end)
 47     {
 48         if(mp[aim]) break;
 49         j = 1;
 50         for(k = str; k <= end; k ++)
 51         {
 52             judge(que[k]);
 53             for(i = 1; i <= 8; i ++)
 54             {
 55                 t[i] = p[9-i];
 56             }
 57             for(i = 1,sum = 0; i <= 8; i ++)
 58                 sum += o[i]*t[i];
 59             if(!mp[sum])
 60             {
 61                 st[end+j] = 1;
 62                 mp[sum] = end+j;
 63                 que[end+j] = sum;
 64                 pre[end+j] = k;
 65                 j ++;
 66             }
 67             for(i = 2; i <= 4; i ++)
 68                 t[i] = p[i-1];
 69             t[1] = p[4];
 70             for(i = 5; i <= 7; i ++)
 71                 t[i] = p[i+1];
 72             t[8] = p[5];
 73             for(i = 1,sum = 0; i <= 8; i ++)
 74                 sum += o[i]*t[i];
 75             if(!mp[sum])
 76             {
 77                 st[end+j] = 2;
 78                 mp[sum] = end+j;
 79                 que[end+j] = sum;
 80                 pre[end+j] = k;
 81                 j ++;
 82             }
 83             for(i = 1; i <= 8; i ++)
 84                 t[i] = p[i];
 85             t[2] = p[7];
 86             t[3] = p[2];
 87             t[6] = p[3];
 88             t[7] = p[6];
 89             for(i = 1,sum = 0; i <= 8; i ++)
 90                 sum += o[i]*t[i];
 91             if(!mp[sum])
 92             {
 93                 st[end+j] = 3;
 94                 mp[sum] = end+j;
 95                 que[end+j] = sum;
 96                 pre[end+j] = k;
 97                 j ++;
 98             }
 99         }
100         str = end+1;
101         end = end+j-1;
102         num ++;
103     }
104     printf("%d\n",num);
105     tail = mp[aim];
106     for(i = num; i >= 1; i --)
107     {
108         ch[i] = st[tail]-1+'A';
109         tail = pre[tail];
110     }
111     for(i = 1; i <= num; i ++)
112     {
113         printf("%c",ch[i]);
114         if(i%60 == 0)
115             printf("\n");
116     }
117     printf("\n");
118     return 0;
119 }
posted @ 2012-11-22 11:13  Naix_x  阅读(193)  评论(0)  编辑  收藏  举报