杭电acm解题报告2001---2099.

				//先dfs搜打掉的球数
				//再dfs2搜会掉下来的球数
				//注意状态保存和球粘连关系就好了
				#include 
				#include 
				using namespace std;
				int H,W,h,w;
				char map[110][110];
				bool v[110][110];
				char color;
				int shoot;
				int d[2][6][2] =
				{
				    {{0,-1},{0,1},{-1,0},{-1,1},{1,0},{1,1}},
				    {{0,-1},{0,1},{-1,0},{-1,-1},{1,0},{1,-1}}
				};
				
				void dfs(int x,int y)
				{ 
				    if(xH)
				        return ;
				    if(yW-(x+1)%2)
				        return ;
				    if(map[x][y-1] == color)
				    {
				        shoot ++;
				        map[x][y-1] = 'E';
				        for (int i=0;i				            dfs( x+d[x%2][i][0], y+d[x%2][i][1] );
				        }
				    }
				}
				
				bool dfs2(int x,int y)
				{
				    char ch;
				
				    if(x				    if(yW-(x+1)%2 || x>H)
				        return false;
				    if(map[x][y-1] == 'E')
				        return false;
				    else
				    {
				        shoot ++;
				        ch = map[x][y-1];
				        map[x][y-1] = 'E';
				        v[x][y-1] = true;
				        for (int i=0;i				            if ( dfs2( x+d[x%2][i][0], y+d[x%2][i][1] ) ) {
				                map[x][y-1] = ch;
				                return true;
				            }
				        }
				    }
				    return false;
				}
				
				int main()
				{
				    int i,j,ans,total;
				    
				    while( scanf("%d %d %d %d",&H,&W,&h,&w)!=EOF )
				    {
				        for(i=1;i				            scanf("%s",map[i]);
				        memset(v,0,sizeof(v));
				        
				        color = map[h][w-1];
				        ans = 0;
				        shoot = 0;
				        if (color != 'E') {
				            dfs(h,w);
				            if(shoot >= 3)    ans += shoot;
				            else {
				                printf("0\n");
				                continue;
				            }
				            for (i=1; i				                for (j=1; j				                    shoot = 0;
				                    if( !v[i][j-1] && map[i][j-1]!='E' && !dfs2(i,j) ) {
				                        ans += shoot;
				                    }
				                }
				            }
				        }
				        printf("%d\n",ans);
				    }
				    return 0;
				}