传送门
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.
Sample Output
3
题意:给你一副图n行m列，其中’W’代表池塘，’.'代表土地，问现在这幅图中有多少个池塘，如果一个池塘的八个方向上如果有池塘，则只算一个,类似于连通块,求有多少个连通块。
思路:bfs和dfs
BFS
用队列实现，需要用到pair或者结构体存储下标，通过找到一个w然后用队列实现连通功能的块
#include<iostream>
#include<queue>
#include<map>
using namespace std;
const int maxn = 105;
char a [maxn][maxn];
pair<int ,int >m1;
typedef pair<int ,int >P;
queue<pair<int ,int > >mo;
void bfs(int x,int y ,int n ,int m){
int i,j;
while(!mo.empty()){
P d = mo.front();
mo.pop();
for(i=-1;i<=1;i++){
for(j=-1;j<=1;j++){
int x1,y1;
x1=d.first+i,y1=d.second+j;
if(x1>=0&&x1<n&&y1>=0&&y1<m&&a[x1][y1]=='W'){
a[x1][y1]='.';
mo.push(P(x1,y1));
}
}
}
}
}
int solve(int n, int m ){
int ans = 0,i,j;
for(i = 0 ;i < n ; i++){
for( j = 0 ;j < m ;j++){
if(a[i][j]=='W'){
ans++;
mo.push(P(i,j));
bfs(i,j,n,m);
}
}
}
return ans;
}
int main()
{
int n , i , j , m ;
cin >> n >> m;
for(i = 0; i < n; i ++){
for(j = 0; j < m ;j++)
cin >> a[i][j];
}
int ans=0;
ans=solve(n,m);
cout << ans <<endl;
}
DFS
查找到了为水坑的低点之后就继续搜索不断查找=。=
#include<iostream>
using namespace std;
const int maxn = 105;
char a [maxn][maxn];
void dfs(int x, int y,int n,int m ){
int i,j,b,c;
for(i = -1;i <= 1;i++){
for(j = -1 ;j <=1 ;j++){
b=i+x;
c=y+j;
if(b>=0&&b<n&&c>=0&&c<m&&a[b][c]=='W'){
a[b][c]='.';
dfs(b,c,n,m);
}
}
}
}
int solve (int n,int m ){
int ans = 0,i,j;
for(i = 0; i < n ;i ++){
for( j = 0 ; j < m ;j ++){
if(a[i][j]=='W'){
dfs(i,j,n,m);
ans++;
}
}
}
return ans;
}
int main()
{
int n , i , j , m ;
cin >> n >> m;
for(i = 0; i < n; i ++){
for(j = 0; j < m ;j++)
cin >> a[i][j];
}
int ans=0;
ans=solve(n,m);
cout << ans <<endl;
}