bzoj1751 [Usaco2005 qua]Lake Counting
2014-07-15 11:52
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1751: [Usaco2005 qua]Lake Counting
Time Limit: 5 Sec Memory Limit: 64 MBSubmit: 168 Solved: 130
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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 figureout 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 12W........WW.
.WWW.....WWW
....WW...WW.
.........WW.
.........W..
..W......W..
.W.W.....WW.
W.W.W.....W.
.W.W......W.
..W.......W.
Sample Output
3OUTPUT DETAILS:
There are three ponds: one in the upper left, one in the lower left,
and one along the right side.
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#include<cstdio> #include<iostream> using namespace std; const int mx[8]={1,1,1,0,0,-1,-1,-1}; const int my[8]={1,0,-1,1,-1,1,0,-1}; int n,m,ans; int map[110][110]; void dfs(int x,int y) { map[x][y]=0; for (int k=0;k<8;k++) { int nx=mx[k]+x; int ny=my[k]+y; if (map[nx][ny]) dfs(nx,ny); } } int main() { scanf("%d%d",&n,&m); for (int i=1;i<=n;i++) for (int j=1;j<=m;j++) { char ch; cin>>ch; if (ch=='W')map[i][j]=1; } for (int i=1;i<=n;i++) for (int j=1;j<=m;j++) if(map[i][j]){dfs(i,j);ans++;} printf("%d",ans); }
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