标签：John .. int Lake WW field POJ Farmer Counting
Lake Counting| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 48370 | Accepted: 23775 |
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
Hint
OUTPUT DETAILS:There are three ponds: one in the upper left, one in the lower left,and one along the right side.
Source
USACO 2004 November【题意】
对于一个图，八个方向代表相邻，求出相邻的（联通）块的个数
【分析】
以一个点W为入口将相邻的W 深搜一遍,同时将他改掉，避免重搜
【代码】
#include<cstdio>
using namespace std;
const int N=105;
int n,m,ans,dir[8][2]={{0,1},{0,-1},{1,0},{-1,0},{-1,-1},{1,-1},{1,1},{-1,1}};
char mp[N][N];
void dfs(int x,int y){
mp[x][y]='.';
for(int i=0;i<8;i++){
int nx=x+dir[i][0];
int ny=y+dir[i][1];
if(nx<1||ny<1||nx>n||ny>m||mp[nx][ny]=='.') continue;
dfs(nx,ny);
}
}
inline void Init(){
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++) scanf("%s",mp[i]+1);
}
inline void Solve(){
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
if(mp[i][j]=='W'){
dfs(i,j);
ans++;
}
}
}
printf("%d",ans);
}
int main(){
Init();
Solve();
return 0;
}
标签：John,..,int,Lake,WW,field,POJ,Farmer,Counting 来源： https://www.cnblogs.com/shenben/p/10353174.html
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