【DFS】【AOJ-61】Lake Counting
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.
Given a diagram of Farmer John's field, determine how many ponds he has.
Input
Line1: Two space-separated integers: N and M
Lines2..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.
Lines2..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
思路:
还是没啥好说的...深度优先搜索 不明白的翻数据结构图那部分
参考代码:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 | #include <stdio.h> #include <string.h> int visited[111][111],movex[8]={-1,-1,0,1,1,1,0,-1},movey[8]={0,-1,-1,-1,0,1,1,1},nx[100000],ny[100000]; char map[111][111]; void dfs( int i, int j); int main() { int n,m; scanf ( "%d%d" ,&n,&m); int i,j; memset (map,-1, sizeof (map)); memset (visited,0, sizeof (visited)); /* for(i=1;i<=n;i++) { for(j=1;j<=m;j++) { char temp; scanf("%c",&temp); if(temp=='.') map[i][j]=0; else map[i][j]=1; } getchar(); } */ memset (map,0, sizeof (map)); for ( int i = 1;i <= n;i ++) scanf ( "%s" ,&map[i][1]); int num=0; for (i=1;i<=n;i++) { for (j=1;j<=m;j++) { if (map[i][j]== 'W' &&!visited[i][j]) { num++; dfs(i,j); } } } printf ( "%d\n" ,num); return 0; } void dfs( int i, int j) { visited[i][j]=1; int k; for (k=0;k<8;k++) { if (map[i+movex[k]][j+movey[k]]== 'W' &&!visited[i+movex[k]][j+movey[k]]) dfs(i+movex[k],j+movey[k]); else continue ; } } |
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