POJ 2386 深度搜索
2016-01-25 16:20
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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
遇到的问题和思路:
这道题目也是书上的例题。我记得我最初看到这道题目的时候,首先是题目没有看懂,不知道什么叫8面连同,在后来看了题目的代码以后才明白是是要遍历周围的8个方位才对。
还有一点就是一定要注意全部都遍历,在dfs中一定要注意这一点,不能忘了。
给出代码:
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 26109 | Accepted: 13115 |
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
遇到的问题和思路:
这道题目也是书上的例题。我记得我最初看到这道题目的时候,首先是题目没有看懂,不知道什么叫8面连同,在后来看了题目的代码以后才明白是是要遍历周围的8个方位才对。
还有一点就是一定要注意全部都遍历,在dfs中一定要注意这一点,不能忘了。
给出代码:
#include<cstdio> #include<cmath> #include<algorithm> #include<cstring> using namespace std; int n , m; char map[105][105]; int ans; int nx , ny; void dfs(int x,int y){ for(int i = -1; i <= 1; i++){ for(int j = -1; j <= 1; j++){ nx = x + i,ny = y + j; if(nx >= 0 && nx < n && ny >= 0 && ny <=m && map[nx][ny] == 'W'){ map[nx][ny] = '.'; dfs(nx , ny); } } } } void solve(){ ans = 0; for(int i = 0; i < n; i++){ for(int j = 0;j < m; j++){ if(map[i][j] == 'W'){ dfs(i , j); ans++; } } } printf("%d\n",ans); } int main(){ while(scanf("%d%d",&n,&m)!=EOF){ memset(map , 0 ,sizeof(map)); for(int i = 0; i < n ; i++){ getchar(); for(int j = 0; j < m ; j++) scanf("%c",&map[i][j]); } solve(); } return 0; }
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