POJ-2386--Lake Counting

原题链接
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

用DFS来做，从有‘W’的地方，将当前位置变成‘.’，把当前d循环遍历8个方向，直到没有W为止

#include <iostream>
#include <cstdio>
using namespace std;
const int N = 120;
char g[N][N];//存储
int n,m;
void dfs(int x,int y){g[x][y] = '.';//将当前位置置为‘.’//循环遍历八个方向for(int dx = -1;dx<=1;dx++){dx//for(int dy = -1;dy<=1;dy++){int nx = x + dx, ny = y + dy;//更新一下//判断当前位置是否越界，是否是W可以继续，如果可以就继续调用if(nx>=0 && nx<=n && ny>=0 && ny<m && g[nx][ny] == 'W')                       dfs(nx,ny);}}return ;
}void solve()
{int res=0;for(int i=0;i<n;i++){for(int j=0;j<m;j++){if(g[i][j]=='W') dfs(i,j),res++;}}printf("%d\n",res);
}
int main()
{std::ios::sync_with_stdio(false);cin.tie(NULL);cin>>n>>m;for(int i=0;i<n;i++){for(int j=0;j<m;j++){cin>>g[i][j];}}solve();return 0;
}

POJ-2386--Lake Counting相关推荐

poj 2386 Lake Counting
poj 2386 Lake Counting 题目链接:http://poj.org/problem?id=2386 题目大意:数湖. 题目 ...
POJ 2386 Lake Counting [DFS]
POJ 2386 Lake Counting 简单的DFS,用了stack代替递归,输入有问题,搞了蛮久,算法是没问题.所以以后一定要记得加上检查输入的那一步然后对于点的定义以后就这么办吧此外还有 ...
POJ 2386 Lake Counting
链接:http://poj.org/problem?id=2386 Lake Counting Time Limit: 1000MS Memory Limit: 65536K Total Submis ...
POJ 2386 Lake Counting DFS水水
http://poj.org/problem?id=2386 题目大意: 有一个大小为N*M的园子,雨后积起了水.八连通的积水被认为是连接在一起的.请求出院子里共有多少水洼? 思路: 水题~直接DFS ...
DFS 之 poj 2386 Lake Counting
// [11/1/2014 JmingS] /* 遍历整个图,找到 'W' 的点,对其周围八个点其进行深搜,若是 'W' 则用 '.' 替换. 最后,在遍历整个图的过程中,找到多少个 'W',即答案. ...
POJ 2386（DFS）
深度优先搜索属于图算法的一种,英文缩写为DFS即Depth First Search.其过程简要来说是对每一个可能的分支路径深入到不能再深入为止,而且每个节点只能访问一次. 举例说明之:下图是一个无向 ...
Lake Counting POJ - 2386
Due to recent rains, water has pooled in various places in Farmer John's field, which is represented ...
DFS:深入优先搜索 POJ-2386 Lake Counting
深度优先搜索是从最开始的状态出发,遍历所有可以到达的状态. 因此可以对所有的状态进行操作,或列举出所有的状态. Lake Counting POJ - 2386 Due to recent rains ...
信息学奥赛一本通（1249：Lake Counting）
1249:Lake Counting 时间限制: 1000 ms 内存限制: 65536 KB 提交数: 9435 通过数: 4902 [题目描述] 题意:有一块N×M的土地, ...
NUC1158 Lake Counting【DFS】
Lake Counting 时间限制: 1000ms 内存限制: 65536KB 通过次数: 1总提交次数: 1 问题描述 Due to recent rains, water has pooled ...

POJ-2386--Lake Counting

POJ-2386--Lake Counting相关推荐

最新文章

热门文章