1 条题解

  • 0
    @ 2025-4-12 21:51:00

    C :

    #include<stdio.h>
    char c[105][105];
    int m,n,dir[8][2]={1,0,-1,0,0,1,0,-1,1,1,1,-1,-1,1,-1,-1};
    void dfs(int x,int y)
    {
    	int i,xx,yy;
    	for(i=0;i<8;i++)
    	{
    		xx=x+dir[i][0];
    		yy=y+dir[i][1];
    		if(c[xx][yy]=='@'&&xx>=0&&yy>=0&&xx<m&&yy<n)
    		{
    			c[xx][yy]='*';
    			dfs(xx,yy);
    		}
    	}
    }
    int main()
    {
    	int i,j,counter;
    	while(scanf("%d%d",&m,&n)!=EOF&&m)
    	{
    		counter=0;
    		for(i=0;i<m;i++)
    		{
    			getchar();
    			scanf("%s",c[i]);
    		}
    		for(i=0;i<m;i++)
    		{
    			for(j=0;j<n;j++)
    			{
    				if(c[i][j]=='@'){counter++;dfs(i,j);}
    			}
    		}
    		printf("%d\n",counter);
    	}
    }
    

    C++ :

    #include<cstdio>
    #include<iostream>
    #include<cstring>
    using namespace std;
    
    char g[100][100];
    int m,n,v[100][100],d[][2]={-1,-1,-1,0,-1,1,0,-1,0,1,1,-1,1,0,1,1};
    
    void dfs(int x,int y)
    {
    	int i,xx,yy;
    	for(i=0;i<8;i++)
    	{
    		xx=x+d[i][0];
    		yy=y+d[i][1];
    		if(xx<0||xx>=m||yy<0||yy>=n||g[xx][yy]=='*'||v[xx][yy])
    			continue;
    		v[xx][yy]=1;
    		dfs(xx,yy);
    	}
    }
    
    int main()
    {
    	int i,j,ans;
    	while(cin>>m>>n,m)
    	{
    		for(i=0;i<m;i++)
    			for(j=0;j<n;j++)
    				cin>>g[i][j];
    		memset(v,0,sizeof(v));
    		for(ans=i=0;i<m;i++)
    			for(j=0;j<n;j++)
    			{
    				if(v[i][j]||g[i][j]=='*')
    					continue;
    				dfs(i,j);
    				ans++;
    			}
    		printf("%d\n",ans);
    	}
    	return 0;
    }
    

    Pascal :

    program OIl_Deposit;
    var i,j,k,n,m,dx,dy,nx,ny,x,y,head,tail,sum:integer;
        gr:array[0..1000,0..1000]of integer;
        dui:array[1..10000,1..2]of integer;
        s:string;
    procedure floodfill(x,y,sum:integer);
    var i,j:integer;
    begin
      dui[head,1]:=x;
      dui[head,2]:=y;
      while head<=tail do
        begin
          for dx:=-1 to 1 do
            begin
              for dy:=-1 to 1 do
                begin
                  nx:=dui[head,1]+dx;
                  ny:=dui[head,2]+dy;
                  if (nx<=m)and(nx>0)and(ny<=n)and(ny>0) then
                    begin
                      if gr[nx,ny]=1 then
                        begin
                          inc(tail);
                          dui[tail,1]:=nx;
                          dui[tail,2]:=ny;
                          gr[nx,ny]:=sum;
                        end;{then}
                    end;{then}
                end;{for-dy}
            end;{for-dx}
          inc(head);
        end;{while}
    end;{floodfill}
    begin{main}
      readln(m,n);
      while m<>0 do
        begin
          for i:=1 to m do
            begin
              readln(s);
              for j:=1 to n do
                begin
                  if s[j]='@' then gr[i,j]:=1;
                  if s[j]='*' then gr[i,j]:=0;
                end;{for-j}
            end;{for-i}
          sum:=10;
          for i:=1 to m do
            begin
              for j:=1 to n do
                begin
                  if gr[i,j]=1 then
                    begin
                      head:=1;
                      tail:=1;
                      inc(sum);
                      floodfill(i,j,sum);
                    end;{then}
                end;{for-j}
            end;{for-i}
          writeln(sum-10);
          readln(m,n);
        end;{while}
    end.
    
    • 1

    信息

    ID
    1980
    时间
    1000ms
    内存
    32MiB
    难度
    (无)
    标签
    递交数
    0
    已通过
    0
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