POJ 3026 Borg Maze(bfs+prim)

题目链接

题意也是挺难懂的。把所有的A,S看成一个点，求这些点的最小生成树。先BFS预处理出来然后prim，注意m,n后边的空格问题，然后数组开小了，也错了次。

  1 #include <cstdio>
  2 #include <cstring>
  3 #include <string>
  4 #include <iostream>
  5 using namespace std;
  6 int p[1001][1001],low[1001];
  7 char str[101][101];
  8 int o[301][301],key[301][301],n,m,kk[1001];
  9 int a[4] = {0,0,1,-1};
 10 int b[4] = {1,-1,0,0};
 11 int quer[20000],quec[20000];
 12 void bfs(int num,int x,int y)
 13 {
 14     int st,end,i,j,step,k;
 15     st = end = 1;
 16     for(i = 0;i < n;i ++)
 17     {
 18         for(j = 0;j < m;j ++)
 19         {
 20             if(str[i][j] == '#')
 21             key[i][j] = 1;
 22             else
 23             key[i][j] = 0;
 24         }
 25     }
 26     quer[1] = x;
 27     quec[1] = y;
 28     p[num][num] = 0;
 29     key[x][y] = 1;
 30     step = 1;
 31     while(st <= end)
 32     {
 33         j = 1;
 34         for(i = st;i <= end;i ++)
 35         {
 36             for(k = 0;k <= 3;k ++)
 37             {
 38                 int r,c;
 39                 r = quer[i]+a[k];
 40                 c = quec[i]+b[k];
 41                 if(r >= 0&&r < n&&c >= 0&&c < m&&!key[r][c])
 42                 {
 43                     key[r][c] = 1;
 44                     if(str[r][c] == 'A'||str[r][c] == 'S')
 45                     p[num][o[r][c]] = step;
 46                     quer[end+j] = r;
 47                     quec[end+j] = c;
 48                     j ++;
 49                 }
 50             }
 51         }
 52         step ++;
 53         st = end+1;
 54         end = end+j-1;
 55     }
 56     return ;
 57 }
 58 int main()
 59 {
 60     int t,i,j,num,ans,mi,z;
 61     char ch[101];
 62     scanf("%d",&t);
 63     while(t--)
 64     {
 65         memset(o,0,sizeof(o));
 66         memset(key,0,sizeof(key));
 67         memset(kk,0,sizeof(kk));
 68         scanf("%d%d",&m,&n);
 69         gets(ch);
 70         for(i = 0;i < n;i ++)
 71         {
 72             gets(str[i]);
 73         }
 74         num = 1;
 75         for(i = 0;i < n;i ++)
 76         {
 77             for(j = 0;j < m;j ++)
 78             {
 79                 if(str[i][j] == 'S'||str[i][j] == 'A')
 80                 o[i][j] = num++;
 81                 else
 82                 key[i][j] = 1;
 83             }
 84         }
 85         num --;
 86         for(i = 0;i < n;i ++)
 87         {
 88             for(j = 0;j < m;j ++)
 89             {
 90                 bfs(o[i][j],i,j);
 91             }
 92         }
 93         ans = 0;
 94         for(i = 1;i <= num;i ++)
 95         low[i] = p[1][i];
 96         kk[1] = 1;
 97         for(i = 1;i <= num-1;i ++)
 98         {
 99             mi = 100000;
100             for(j = 1;j <= num;j ++)
101             {
102                 if(mi > low[j]&&!kk[j])
103                 {
104                     mi = low[j];
105                     z = j;
106                 }
107             }
108             kk[z] = 1;
109             ans += mi;
110             for(j = 1;j <= num;j ++)
111             {
112                 if(low[j] > p[z][j]&&!kk[j])
113                 low[j] = p[z][j];
114             }
115         }
116         printf("%d\n",ans);
117     }
118     return 0;
119 }

 

posted @ 2012-12-08 11:19  Naix_x  阅读(148)  评论(0)  编辑  收藏  举报