pku1979

 

Source Code

Problem: 1979		User: henry11
Memory: 356K		Time: 0MS
Language: G++		Result: Accepted

• Source Code

• /*
  廣度搜索 
  queue 為訪問堆； 
  status 表示訪問狀態；
  map 表示地圖； 
  */
  
  #include<stdio.h>
  #include<string.h>
  
  struct node
  {
      int a, b;
  };
  
  int direct[4][2] = {-1, 0, 0, 1, 1, 0, 0, -1};
  node queue[401];
  char map[21][21];
  int head, tail;
  int count, len, height;
  
  void initQueue(int x, int y)
  {
      queue[0].a = x;
      queue[0].b = y;
      map[x][y] = '*';
      head = tail = 0;
  }         
  
  void push(int x, int y)
  {
      queue[tail+1].a = x;
      queue[tail+1].b = y;
      map[x][y] = '*';
      tail++;
  }   
  
  void DFS(int prea, int preb)
  {
      int n;
      int i, ta = prea, tb = preb;
      initQueue(ta, tb);
      //printf("%d %d/n", queue[head].a, queue[head].b);
      count = 1;
      while(head <= tail)
      {
          for(i=0; i<4; i++)
          {
              ta = queue[head].a+direct[i][0];
              tb = queue[head].b+direct[i][1];
              //printf("ta=%d+%d,tb=%d+%d/n", queue[head].a, direct[i][0],queue[head].b, direct[i][1]);
              //printf("ta=%d tb=%d/n", ta, tb);
              if(ta>=0 && ta<height && tb>=0 && tb<len)
              {
                  //printf("%c/n", map[ta][tb]);
                  if(map[ta][tb] == '.')
                  {
                      push(ta, tb);
                      count++;
                  }    
              }
              //printf("%d %d %d/n", i, queue[head].a, queue[head].b);    
          }
          //scanf("%d", &n);
          head++;    
      } 
      printf("%d/n", count);   
  }    
  
  int main()
  {
      int i, j;
      int pa, pb;
      while(scanf("%d%d", &len, &height) != EOF)
      {
          //printf("%d %d/n", len, height);
          if(!len && !height)break;
          for(i=0; i<height; i++)
          {
              scanf("%s", map[i]);
              for(j=0; j<len; j++)
              {
                  if(map[i][j] == '@')
                  {
                      pa = i;
                      pb = j;
                  }
              }      
          }
          //printf("%d %d/n", pa, pb);
          DFS(pa, pb);  
      }    
      return 1;
  }     

又一道寬搜的題目，練得還可以，深搜的話是不是必須就得遞歸了。