POJ 2396 Lake Counting(简单dfs)
2016-03-29 15:25
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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.
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
Sample Output
这个题目是简单的深度搜索题目,题意:连天下雨,农夫的n*m田地里有积水,积水连成了小水洼,让你计算小水洼有几个。
易知,这个题目的做法是dfs,因为其“从某一个状态转移到另一个状态,直到其无法转移为止,且无固定初末状态”。
代码如下:
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 26942 | Accepted: 13532 |
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
这个题目是简单的深度搜索题目,题意:连天下雨,农夫的n*m田地里有积水,积水连成了小水洼,让你计算小水洼有几个。
易知,这个题目的做法是dfs,因为其“从某一个状态转移到另一个状态,直到其无法转移为止,且无固定初末状态”。
代码如下:
#include<iostream> using namespace std; int n,m; char str[105][105]; void solve(); void dfs(int ,int ); int main() { cin>>n>>m;//n行m列 for(int i=0;i<n;i++) cin>>str[i]; solve(); return 0; } void solve() { int res=0; for(int i=0;i<n;i++) for(int j=0;j<m;j++) if(str[i][j]=='W') { dfs(i,j); res++; } cout<<res<<endl; } void dfs(int x,int y) { int nx,ny; str[x][y]='.';//走过的水坑更新为陆地 //遍历上下左右八个方向 for(int fx=-1;fx<=1;fx++) for(int fy=-1;fy<=1;fy++) { nx=x+fx,ny=y+fy; //判断遍历的点是否在n*m的地盘内以及是否是水坑 if(nx>=0&&nx<n&&ny>=0&&ny<m&&str[nx][ny]=='W') dfs(nx,ny); } }
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