【BFS】poj 2386 Lake Counting

Lake Counting

Time Limit: 1000MS
Memory Limit: 65536KTotal Submissions: 35654
Accepted: 17691

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 12W........WW..WWW.....WWW....WW...WW..........WW..........W....W......W...W.W.....WW.W.W.W.....W..W.W......W...W.......W.

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.

///AC代码

#include <iostream>#include <set>#include <map>#include <stack>#include <cmath>#include <queue>#include <cstdio>#include <bitset>#include <string>#include <vector>#include <iomanip>#include <cstring>#include <algorithm>#include <functional>#define PI acos(-1)#define eps 1e-8#define inf 0x3f3f3f3f#define debug(x) cout<<"---"<<x<<"---"<<endltypedef long long ll;using namespace std;int m, n;char str[105][105];bool vis[105][105];struct node{    int x, y;};int dix[8] = {1, -1, 0, 0, 1, -1, -1, 1};int diy[8] = {0, 0, 1, -1, 1, -1, 1, -1};int cango(int x, int y){    return x >= 0 && x < m && y >= 0 && y < n && vis[x][y] == false;}int bfs(int x, int y){    //pp++;    queue<node> qu;    node gg;    gg.x = x;    gg.y = y;    qu.push(gg);    memset(vis, false, sizeof(vis));    vis[x][y] = true;    str[x][y] = '.';    int sum = 0;    while (!qu.empty())    {        gg = qu.front();        qu.pop();        sum++;        for (int i = 0; i < 8; i++)        {            int _x = gg.x + dix[i];            int _y = gg.y + diy[i];            if (cango(_x, _y) && str[_x][_y] == 'W')            {                node hh;                hh.x = _x;                hh.y = _y;                qu.push(hh);                vis[_x][_y] = true;                str[_x][_y] = '.';            }        }    }    return sum;}int main(){    while (cin >> m >> n)    {        //memset(vis, false, sizeof(vis));        int ans = 0, pp = 0;        for (int i = 0; i < m; i++)        {            cin >> str[i];        }        for (int i = 0; i < m; i++)        {            for (int j = 0; j < n; j++)            {                if (str[i][j] == 'W')                {                    ans += bfs(i, j);                    pp++;                }            }        }        //cout << ans << endl;        cout << pp << endl;    }    return 0;}