http://blog.csdn.net/c20182030/article/details/52327948
1388:Lake Counting
- 总时间限制:
- 1000ms 内存限制:
- 65536kB
- 描述
- 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. 输入
- * 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. 输出
- * Line 1: The number of ponds in Farmer John's field. 样例输入
-
10 12W........WW..WWW.....WWW....WW...WW..........WW..........W....W......W...W.W.....WW.W.W.W.....W..W.W......W...W.......W.
样例输出 -
3
提示 - OUTPUT DETAILS: There are three ponds: one in the upper left, one in the lower left,and one along the right side. 来源
- USACO 2004 Novembe
- 题目大意
- 大雨过后，农夫John的农场
被水淹没不知所措有了积水，农场地图中‘W’代表积水，'.'代表干燥的土地。八连通的积水在同一个水洼里。农夫John想知道农场里有多少水洼。 - 这是一道搜索题。与其他搜索题一样，都是不断地搜索，直到达到目的。但是，这道题的目的是什么呢？
- 目的是求出有多少水洼，如果递归的话，怎么才能知道求出了数量呢？
- 所以这样，不能单纯的只调用一次函数（主程序中），否则最多只会找出一个水洼。
- 我一开始是这样想的，把每个水洼只留下一个W，其余换成干燥，最后再看有几个W。可是这样搜索的时候，留下的W也会被搜到，搜到就会变成'.'，答案就会是0。为了避免这个问题，需要设立一个辅助数组或者把这个W换成另一个字符，但是这样既浪费时间也浪费空间（还浪费我的脑容量）。
- 于是我把函数写在循环里，把一个水洼消灭干净后，水洼数量就加1。我用DFS，BFS实现了这个想法。