DFS模板题---Lake Counting
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Lake Counting
Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
Submit
Status
Practice
POJ 2386
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
Sample Output
3
Hint
OUTPUT DETAILS:
There are three ponds: one in the upper left, one in the lower left,and one along the right side.
标准的DFS模板题目,代码如下:
#include <stdio.h>#include <string.h>#define lim 500void dfs(int x, int y, int map[][lim], int vis[][lim]);int main (){ int row, col; while(scanf("%d%d", &row, &col)!=EOF) { if(row==0) return 0; int i, j, count=0; char str[800]; int map[500][500], vis[500][500];//定义两个数组 memset(str, 0, sizeof(str)); memset(map, 0, sizeof(map)); memset(vis, 0, sizeof(vis)); for(i=0;i<row;i++)//进行题目中的输入 { scanf("%s", str);//输入方法 for(j=0;j<col;j++) map[i+1][j+1]=str[j]-'.';//相当于外边加一层 * ,防止搜索时数组越界!!! } for(i=1;i<=row;i++) for(j=1;j<=col;j++) if(map[i][j]==41 && vis[i][j]==0)//搜索时遇到还未被访问的@则进行搜索 { dfs(i, j, map, vis); count++; } printf("%d\n", count); } return 0;}void dfs(int x, int y, int map[][lim], int vis[][lim]){ if(map[x][y]==0||vis[x][y]==1) return; vis[x][y]=1; dfs(x, y-1, map, vis );//上 dfs(x, y+1, map, vis );//下 dfs(x-1, y , map, vis);//左 dfs(x+1, y, map, vis);//右 dfs(x-1, y-1, map, vis);//左上 dfs(x-1, y+1, map, vis );//左下 dfs(x+1, y-1, map, vis );//右上 dfs(x+1, y+1, map, vis );//右下}
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