dfs/poj 2386 Lake Counting

 1 #include<cstdio>
 2 using namespace std;
 3 const int b[8][2]={{-1,-1},{-1,0},{-1,1},{1,-1},{1,0},{1,1},{0,-1},{0,1}};
 4 int n,m;
 5 char a[110][110];
 6
 7 void dfs(int x,int y)
 8 {
 9     a[x][y]=‘.‘;
10     for (int k=0;k<8;k++)
11     {
12         int dx=x+b[k][0];
13         int dy=y+b[k][1];
14         if (0<=dx && dx<n && 0<=dy && dy<m && a[dx][dy]==‘W‘) dfs(dx,dy);
15     }
16     return;
17 }
18
19 int main()
20 {
21     scanf("%d%d",&n,&m);
22     for (int i=0;i<n;i++) scanf("%s",a[i]);
23
24     int ans=0;
25     for (int i=0;i<n;i++)
26         for (int j=0;j<m;j++)
27             if (a[i][j]==‘W‘)
28             {
29                 dfs(i,j);
30                 ans++;
31             }
32     printf("%d\n",ans);
33     return 0;
34 }
时间: 2024-11-02 12:46:04

dfs/poj 2386 Lake Counting的相关文章

POJ 2386 Lake Counting 搜索题解

简单的深度搜索就可以了,看见有人说什么使用并查集,那简直是大算法小用了. 因为可以深搜而不用回溯,故此效率就是O(N*M)了. 技巧就是增加一个标志P,每次搜索到池塘,即有W字母,那么就认为搜索到一个池塘了,P值为真. 搜索过的池塘不要重复搜索,故此,每次走过的池塘都改成其他字母,如'@',或者'#',随便一个都可以. 然后8个方向搜索. #include <stdio.h> #include <vector> #include <string.h> #include

《挑战》2.1 POJ 2386 Lake Counting (简单的dfs)

1 # include<cstdio> 2 # include<iostream> 3 4 using namespace std; 5 6 # define MAX 123 7 8 char grid[MAX][MAX]; 9 int nxt[8][2] = { {1,0},{0,-1},{-1,0},{0,1},{1,1},{1,-1},{-1,1},{-1,-1} }; 10 int n,m; 11 12 int can_move( int x,int y ) 13 { 14

poj 2386 Lake Counting

Lake Counting Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 24578   Accepted: 12407 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 <= 10

poj 2386 Lake Counting(BFS解法)

Lake Counting Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 45142   Accepted: 22306 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 <= 10

poj - 2386 Lake Counting &amp;&amp; hdoj -1241Oil Deposits (简单dfs)

http://poj.org/problem?id=2386 http://acm.hdu.edu.cn/showproblem.php?pid=1241 求有多少个连通子图.复杂度都是O(n*m). 1 #include <cstdio> 2 3 char filed[110][110]; 4 int n,m; 5 void dfs(int x,int y) 6 { 7 for(int i=-1;i<=1;i++) 8 for(int j=-1;j<=1;j++) //循环遍历8

POJ 2386 Lake Counting (水题,DFS)

题意:给定一个n*m的矩阵,让你判断有多少个连通块. 析:用DFS搜一下即可. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring&

POJ 2386 lake counting DFS

朴素DFS 计算有多少个水坑 TEST 10 12 W........WW. .WWW.....WWW ....WW...WW. .........WW. .........W.. ..W......W.. .W.W.....WW. W.W.W.....W. .W.W......W. ..W.......W. 3 /* DFS,挑战程序设计竞赛 巫 */ #include<cstdio> #include<iostream> using namespace std; string

POJ 2386 Lake Counting(DFS)

题意: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在一起的.求园子里一共有多少水洼? * * * * W*    (八连通指的就是左图中相对W的*的部分) * * * 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 O

POJ 2386 - Lake Counting 题解

此文为博主原创题解,转载时请通知博主,并把原文链接放在正文醒目位置. 题目链接:http://poj.org/problem?id=2386 题目大意: 给出一张N*M的地图(1<=N,M<=100),地图上的W表示水坑,.表示陆地.水坑是八方向连通的,问图中一共有多少个水坑. Sample Input 9 5 2 1 5 2 1 5 2 1 4 1 2 3 4 0 Sample Output 6 5 分析: 非常简单的深搜.两重循环遍历这个地图,遇到W就dfs来求出这一整个大水坑,并把搜索过