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, 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

Hint

OUTPUT DETAILS:
There are three ponds: one in the upper left, one in the lower left,and one along the right side.

#include<iostream>
#include<string>
#include<cstdio>
#include<cstring>
#include<queue>
#include<map>
#include<set>
#include<algorithm>
#include<cmath>
#include<cstdlib>
using namespace std;int m, n;
char garden[105][105];void dfs(int x, int y)
{//将当前点取消标记，避免重复查找garden[x][y] = '.';//遍历周围的八个点for (int dx = -1; dx <= 1; dx++){for (int dy = -1; dy <= 1;dy++){int nx = x + dx;int ny = y + dy;if(0<=nx && nx<n && 0<=ny && ny<m && garden[nx][ny]=='W')dfs(nx, ny);}}
}int main(void)
{while(~scanf("%d%d", &n,&m)){getchar();  //吸收两数字后的换行符memset(garden, 0, sizeof(garden));for (int i = 0; i < n;i++){for (int j = 0; j < m;j++)scanf("%c", &garden[i][j]);getchar();  //吸收每次输入一行后的换行符}int sum = 0;for (int i = 0; i < n; i++){for (int j = 0; j < m; j++){if (garden[i][j] == 'W'){dfs(i, j);sum++;}}}cout << sum << endl;}return 0;
}

DFS:深入优先搜索 POJ-2386 Lake Counting相关推荐

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
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
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 ...
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 ...
Openjudge1388 Lake Counting【DFS/Flood Fill】
http://blog.csdn.net/c20182030/article/details/52327948 1388:Lake Counting 总时间限制: 1000ms 内存限制: 65 ...

DFS:深入优先搜索 POJ-2386 Lake Counting

Lake Counting

DFS:深入优先搜索 POJ-2386 Lake Counting相关推荐

最新文章

热门文章