1751: [Usaco2005 qua]Lake Counting

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 190  Solved: 150
[Submit][Status][Discuss]

Description

Due to recent rains, water has pooled in various places in Farmer John's field, which is represented by a rectangle of N x M (1 <= N <= 100; 1 <= M <= 100) squares. Each square contains either water ('W') or dry land ('.'). Farmer John would like to figure out how many ponds have formed in his field. A pond is a connected set of squares with water in them, where a square is considered adjacent to all eight of its neighbors. Given a diagram of Farmer John's field, determine how many ponds he has.

Input

* Line 1: Two space-separated integers: N and M * Lines 2..N+1: M characters per line representing one row of Farmer John's field. Each character is either 'W' or '.'. The characters do not have spaces between them.

Output

* Line 1: The number of ponds in Farmer John's field.

Sample Input

10 12
W........WW.
.WWW.....WWW
....WW...WW.
.........WW.
.........W..
..W......W..
.W.W.....WW.
W.W.W.....W.
.W.W......W.
..W.......W.

Sample Output

3

OUTPUT DETAILS:

There are three ponds: one in the upper left, one in the lower left,
and one along the right side.

HINT

 

Source

Gold

题解:直接萌萌哒DFS秒之,经典的普及组难度基础题,水水哒

(Tip:38行的dfs(a1,a2)貌似只有这样写在本机才能对,提交也能A;很神奇的是如果直接写dfs(i,j)的话在本机就会出现带入的是(1,1)结果进去的是(2,2)QAQ,然后各种神奇跪OTL,更神奇的是这个在本机都跪成狗的程序居然submit之后也能A(QAQ),求神犇解释)

 var
i,j,k,l,m,n,a1,a2:longint;
c1:char;
a:array[..,..] of longint;
procedure dfs(x,y:longint);inline;
begin
a[x,y]:=;
if a[x-,y-]= then dfs(x-,y-);
if a[x,y-]= then dfs(x,y-);
if a[x+,y-]= then dfs(x+,y-);
if a[x-,y+]= then dfs(x-,y+);
if a[x,y+]= then dfs(x,y+);
if a[x+,y+]= then dfs(x+,y+);
if a[x-,y]= then dfs(x-,y);
if a[x+,y]= then dfs(x+,y);
end;
begin
readln(n,m);
fillchar(a,sizeof(a),);
for i:= to n do
begin
for j:= to m do
begin
read(c1);
case c1 of
'W':a[i,j]:=;
'.':a[i,j]:=;
end;
end;
readln;
end;
l:=;
for i:= to n do
for j:= to m do
if a[i,j]= then
begin
a1:=i;a2:=j;
inc(l);dfs(a1,a2);
end;
writeln(l);
readln;
end.

1751: [Usaco2005 qua]Lake Counting的更多相关文章

  1. bzoj1751 [Usaco2005 qua]Lake Counting

    1751: [Usaco2005 qua]Lake Counting Time Limit: 5 Sec  Memory Limit: 64 MB Submit: 168  Solved: 130 [ ...

  2. POJ_2386 Lake Counting (dfs 错了一个负号找了一上午)

    来之不易的2017第一发ac http://poj.org/problem?id=2386 Lake Counting Time Limit: 1000MS   Memory Limit: 65536 ...

  3. POJ 2386 Lake Counting(深搜)

    Lake Counting Time Limit: 1000MS     Memory Limit: 65536K Total Submissions: 17917     Accepted: 906 ...

  4. POJ 2386 Lake Counting

    Lake Counting Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 28966   Accepted: 14505 D ...

  5. BZOJ1754: [Usaco2005 qua]Bull Math

    1754: [Usaco2005 qua]Bull Math Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 374  Solved: 227[Submit ...

  6. BZOJ2023: [Usaco2005 Nov]Ant Counting 数蚂蚁

    2023: [Usaco2005 Nov]Ant Counting 数蚂蚁 Time Limit: 4 Sec  Memory Limit: 64 MBSubmit: 56  Solved: 16[S ...

  7. BZOJ 3385: [Usaco2004 Nov]Lake Counting 数池塘

    题目 3385: [Usaco2004 Nov]Lake Counting 数池塘 Time Limit: 1 Sec  Memory Limit: 128 MB Description     农夫 ...

  8. 1755: [Usaco2005 qua]Bank Interest

    1755: [Usaco2005 qua]Bank Interest Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 187  Solved: 162[Su ...

  9. 1753: [Usaco2005 qua]Who's in the Middle

    1753: [Usaco2005 qua]Who's in the Middle Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 290  Solved:  ...

随机推荐

  1. 实例:基于ListActivity实现列表

    如果程序的窗口仅仅需要显示一个列表,则可以直接让Activity继承ListActivity来实现,ListActivity的子类无须调用setContentView()方法来显示某个界面,而是可以直 ...

  2. Raphael的text及对齐方式

    Raphael的text及对齐方式 <%@ page language="java" contentType="text/html; charset=UTF-8&q ...

  3. CodeForces 512B(区间dp)

    D - Fox And Jumping Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64 ...

  4. Redis 提供的好的解决方案 实例

    1.Redis 支持的String 类型的,它是 二进制的,比较安全,当有一些 图片或者css文件影响运行速度的时候,可以缓存之.

  5. 最快让你上手ReactiveCocoa之基础篇(简称RAC)

    前言 很多blog都说ReactiveCocoa好用,然后各种秀自己如何灵活运用ReactiveCocoa,但是感觉真正缺少的是一篇如何学习ReactiveCocoa的文章,小编看了很多篇都没看出怎么 ...

  6. TCP&UDP

    TCP(传输控制协议) 建立连接,形成传输数据的通道 在连接中进行大数据传输(数据大小不受限制) 通过三次握手完成连接,是可靠协议,安全送达(三次握手向服务器发送请求,响应请求回复,发送数据) 必须建 ...

  7. C# OpenFileDialog 使用

    OpenFileDialog ofd = new OpenFileDialog(); //设置标题 ofd.Title = "选择文件"; //是否保存上次打开文件的位置 ofd. ...

  8. Function.caller、arguments.caller、argument.callee

    caller.callee是与javascript函数相关的两个属性,今天来总结下. Function.caller caller是javascript函数的一个属性,它指向调用当前函数的函数,如果函 ...

  9. Surface Dial 与 Windows Wheel UWP应用开发

    随着微软发布 Surface Studio 在演示视频中非常抢眼的一个配件就是 Surface Dial,Dial 是Windows输入设备大家庭中的新成员我们把它归类为Windows Wheel 类 ...

  10. 解决 Eclipse build workspace validation javascript 慢的问题

    参考: http://blog.csdn.net/zhangzikui/article/details/24805935 http://www.cnblogs.com/wql025/p/4978351 ...