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  • poj3083Children of the Candy Corn(dfs+bfs)

    http://poj.org/problem?id=3083纠结了好久 被它的转向搞晕 了

    WA两次 加一else AC。。

    设置四个方向 初始化方向自己算出来 初始化在数组中 dfs+bfs

    左  就是当前方向向左 左走不动就逆时针走下一个方向  依次类推 右同样的方式

    View Code
      1 #include <iostream>
      2  #include<cstdio>
      3  #include<cstring>
      4  using namespace std;
      5  char s[50][50],so[2];
      6  int pr[7],dis[4][2]={-1,0,0,-1,1,0,0,1},vis[50][50],n,m,flag;
      7  int dir[2][4][4] = {{{1, 0, 3, 2}, {2, 1, 0, 3}, {3, 2, 1, 0}, {0, 3, 2, 1}},{{3, 0, 1, 2}, {0, 1, 2, 3}, {1, 2, 3, 0}, {2, 3, 0, 1}}};
      8  struct node
      9  {
     10      int x,y,num;
     11  }q[100010];
     12  int judge(int x,int y)
     13  {
     14      if(x<1||x>n||y<1||y>m)
     15      return 0;
     16      if(s[x][y]=='#')
     17      return 0;
     18      return 1;
     19  }
     20  void bfs()
     21  {
     22      int i,j,p = 0, d = 1;
     23      memset(vis,0,sizeof(vis));
     24      q[d].x = so[0];
     25      q[d].y = so[1];
     26      q[d].num = 1;
     27      vis[so[0]][so[1]] = 1;
     28      while(p!=d)
     29      {
     30          p++;
     31          int tx = q[p].x;
     32          int ty = q[p].y;
     33          int tnum = q[p].num;
     34          if(s[tx][ty]=='E')
     35          {
     36              pr[5] = tnum;
     37              break;
     38          }
     39          for(i = 0 ; i < 4 ; i++)
     40          {
     41              int nx = tx+dis[i][0];
     42              int ny = ty+dis[i][1];
     43              if(!vis[nx][ny]&&judge(nx,ny))
     44              {
     45                  vis[nx][ny] = 1;
     46                  d++;
     47                  q[d].x = nx;
     48                  q[d].y = ny;
     49                  q[d].num = tnum+1;
     50              }
     51          }
     52      }
     53  }
     54  void dfs(int x,int y,int d,int f)
     55  {
     56      pr[f]++;
     57      if(s[x][y]=='E')
     58          return ;
     59      int i,j,td,tx,ty;
     60      td = dir[f][d][0];
     61      tx = x+dis[td][0];
     62      ty = y+dis[td][1];
     63      if(judge(tx,ty))
     64            dfs(tx,ty,td,f);
     65      else
     66      {
     67          td = dir[f][d][1];
     68          tx = x+dis[td][0];
     69          ty = y+dis[td][1];
     70          if(judge(tx,ty))
     71          dfs(tx,ty,td,f);
     72          else
     73          {
     74              td = dir[f][d][2];
     75              tx = x+dis[td][0];
     76              ty = y+dis[td][1];
     77              if(judge(tx,ty))
     78              dfs(tx,ty,td,f);
     79              else
     80              {
     81                  td = dir[f][d][3];
     82                  tx = x+dis[td][0];
     83                  ty = y+dis[td][1];
     84                  if(judge(tx,ty))
     85                  dfs(tx,ty,td,f);
     86              }
     87          }
     88      }
     89  }
     90  int main()
     91  {
     92      int i,j,k,t,in;
     93      cin>>t;
     94      while(t--)
     95      {
     96          cin>>m>>n;
     97          pr[0] = 0;
     98          pr[1] = 0;
     99          for(i = 1 ; i <= n ; i++)
    100          {
    101              getchar();
    102              for(j = 1 ; j <= m ; j++)
    103              {
    104                  cin>>s[i][j];
    105                  if(s[i][j]=='S')
    106                  {
    107                      so[0] = i;
    108                      so[1] = j;
    109                      if(so[0]==n)
    110                      in = 0;
    111                      if(so[0]==1)
    112                      in = 2;
    113                      if(so[1]==m)
    114                      in = 1;
    115                      if(so[1]==1)
    116                      in = 3;
    117                  }
    118              }
    119          }
    120          bfs();
    121          for(i = 0 ; i < 2 ; i++)
    122          {
    123              memset(vis,0,sizeof(vis));
    124              vis[so[0]][so[1]] = 1;
    125              dfs(so[0],so[1],in,i);
    126          }
    127          cout<<pr[0]<<" "<<pr[1]<<" "<<pr[5]<<endl;
    128      }
    129      return 0;
    130  }
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  • 原文地址:https://www.cnblogs.com/shangyu/p/2874121.html
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