大家好,又见面了,我是全栈君,今天给大家准备了Idea注册码。
转载请注明出处:题目链接:http://poj.org/problem?id=1562
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Description
Given a diagram of Farmer John’s field, determine how many ponds he has.
Input
* 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
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
3
代码例如以下:
#include <iostream> #include <algorithm> using namespace std; #include <cstring> #define TM 100+17 int N, M; char map[TM][TM]; bool vis[TM][TM]; int xx[8]={0,1,1,1,0,-1,-1,-1}; int yy[8]={1,1,0,-1,-1,-1,0,1}; void DFS(int x, int y) { vis[x][y] = true; for(int i = 0; i < 8; i++) { int dx = x+xx[i]; int dy = y+yy[i]; if(dx>=0&&dx<N&&dy>=0&&dy<M&&!vis[dx][dy]&&map[dx][dy] == 'W') { vis[dx][dy] = true; DFS(dx,dy); } } } int main() { int i, j; while(cin>>N>>M) { int count = 0; memset(vis,false,sizeof(vis)); for(i = 0; i< N; i++) { cin>>map[i]; } for(i = 0; i < N; i++) { for(j = 0; j < M; j++) { if(map[i][j] == 'W' && !vis[i][j]) { count++; DFS(i,j); } } } cout<<count<<endl; } return 0; }
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