題目描述
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.
輸入描述:
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.
輸出描述:
Line 1: The number of ponds in Farmer John’s field.
示例1
輸入
10 12
W........WW.
.WWW.....WWW
....WW...WW.
.........WW.
.........W..
..W......W..
.W.W.....WW.
W.W.W.....W.
.W.W......W.
..W.......W.
輸出
3
解題思路
深搜每個點位，並把他 ‘W’ 填爲 ‘.’，這樣每遇一次水，計數一次，且將附近的所有水填掉，全局遍歷解決，此題遍歷每個點不是複雜度的關鍵，減少不必要的探索水關鍵，並非多好思路，您若有更好思路，敬請評論或私信。
#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
using namespace std;
int n,m;
void Dfs(vector<string> &sv,int x,int y)
{
if(x >= 0 && x < n && y >= 0 && y < m && sv[x][y] == 'W')
{
sv[x][y] = '.';
for(int i = -1;i < 2;++i)
{
for(int j = -1;j < 2;++j)
{
if(!i && !j)
continue;
Dfs(sv,x + i,y + j);
}
}
}
}
int main()
{
while(cin >> n >> m)
{
int count = 0;
vector<string> sv(n);
for(auto &e : sv)
cin >> e;
for(int i = 0;i < n;++i)
{
for(int j = 0;j < m;++j)
{
if(sv[i][j] == 'W')
{
Dfs(sv,i,j);
++count;
}
}
}
cout << count << endl;
}
return 0;
}