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 <= 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.
/**********************
author: yomi
date: 18.8.14
ps:现在的行为大概像以卵击石，没办法，死也要挣扎着死。
**********************/
#include <iostream>
#include <queue>
#include <cstdio>
using namespace std;
const int maxn = 150;
bool vis[maxn][maxn];
int m, n;
char g[maxn][maxn];
struct Pos
{
int x, y;
}p[maxn];
int dx[3] = {-1, 0, 1};
int dy[3] = {-1, 0, 1};
void bfs(int x, int y)
{
queue<Pos>q;
while(!q.empty()){
q.pop();
}
Pos p;
p.x = x;
p.y = y;
q.push(p);
vis[x][y] = true;
while(!q.empty()){
Pos now = q.front();
q.pop();
for(int i=0; i<3; i++){
for(int j=0; j<3; j++){
int newX = now.x + dx[i];
int newY = now.y + dy[j];
if(newX<0||newX>=n||newY<0||newY>=m)
continue;
if(g[newX][newY]==‘.‘)
continue;
if(!vis[newX][newY]){
Pos New;
New.x = newX;
New.y = newY;
q.push(New);
vis[newX][newY] = true;
}
}
}
}
}
int main()
{
cin >> n >> m;
int ans = 0;
for(int i=0; i<n; i++){
scanf("%s", g[i]);
}
for(int i=0; i<n; i++){
for(int j=0; j<m; j++){
if(g[i][j] == ‘W‘ && !vis[i][j]){
bfs(i, j);
ans++;
}
}
}
cout << ans;
return 0;
}
/**
10 12
W........WW.
.WWW.....WWW
....WW...WW.
.........WW.
.........W..
..W......W..
.W.W.....WW.
W.W.W.....W.
.W.W......W.
..W.......W.
**/
原文地址：https://www.cnblogs.com/AbsolutelyPerfect/p/9476773.html