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.
3
题目大意： W表示水域，可以连接8个方向，要求你求出约翰的田里能形成几个水域
思路：这是一个典型的搜索题目，可以考虑深搜和广搜。
//1009 Lake Counting 有 DFS写的 //深搜
#include <stdio.h>
#include <string.h>
const int maxn = 100 + 5;
int dx[8]= {1,1,1,-1,-1,-1,0,0};
int dy[8]= {1,0,-1,1,0,-1,1,-1};
//定义八个方向
char a[maxn][maxn]; //存储地图,以及标记该点能否再访问
int n, m;
void dfs(int x, int y)
{
a[x][y] = ‘.‘; //标记已访问过。
for(int i=0; i<8; ++i)
{
int tx = x + dx[i];
int ty = y + dy[i];
if( tx>=0&&tx<n&&ty>=0&&ty<m&&a[tx][ty]==‘W‘)
//如果(tx,ty)位置是合法的，则搜索下去
{
dfs(tx,ty);
}
}
}
int main()
{
int i, j;
while(scanf("%d%d",&n,&m)!=EOF)
{
for(i=0; i<n; ++i) scanf("%s",a[i]);
int total = 0;
for(i=0; i<n; ++i)
for(j=0; j<m; ++j)
{
if(a[i][j]==‘W‘)
{
//只要找到了一个‘w‘则对其所在的块进行标记
total++;//结果加一
dfs(i,j);
}
}
printf("%d\n",total);
}
return 0;
}
//-------------------------------------------
//BFS, 用数组实现队列 //广搜
#include <stdio.h>
#include <string.h>
const int maxn = 105;//地图大小
char Map[maxn][maxn];
int visit[maxn][maxn];
int n, m;
const int queue_size = 10000;//队列大小
struct node
{
int x, y;
};
node q[queue_size];
int front, rear; //队首指针和队尾指针
int dx[8]= {1,1,1,-1,-1,-1,0,0};
int dy[8]= {1,0,-1,1,0,-1,1,-1};
void bfs(int x, int y)
{
int i, tx, ty;
node t, next;
front = rear = 0;
t.x = x;
t.y = y;
q[front] = t; //起点入队
visit[x][y] = 1; //将起点标记为已访问
while(front<=rear) //如果队列不为空
{
t = q[front]; front++; //拿出队首元素
for(i=0; i<8; ++i)
{
tx = t.x + dx[i];
ty = t.y + dy[i];
if( tx>=0&&tx<n&&ty>=0&&ty<m&&Map[tx][ty]==‘W‘&&visit[tx][ty]==0)
//如果(tx,ty)位置合法，则加入队列
{
next.x = tx, next.y = ty;
q[++rear] = next;
//并且将其标记为已访问
Map[next.x][next.y] = ‘.‘;
}
}
}
}
int main()
{
int i, j, total;
while(scanf("%d%d",&n,&m)!=EOF)
{
for(i=0; i<n; ++i) scanf("%s",Map[i]);
total = 0;
memset( visit, 0, sizeof visit );
for(i=0; i<n; ++i)
{
for(j=0; j<m; ++j)
{
if(Map[i][j]==‘W‘ && visit[i][j] == 0)
{
total++;
bfs(i,j);
}
}
}
printf("%d\n",total);
}
return 0;
}
//-------------------------------------------
//1009 Lake Counting 用BFS，队列使用C++_STL中的封装的queue，包含在头文件#include<queue>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
const int maxn = 105;//地图大小
char Map[maxn][maxn];
int n, m;
int dx[8]= {1,1,1,-1,-1,-1,0,0};
int dy[8]= {1,0,-1,1,0,-1,1,-1};
struct node
{
int x, y;
};
void bfs(int x, int y)
{
queue<node> q;
int i, tx, ty;
node t, next;
t.x = x; t.y = y;
q.push(t);
Map[x][y] = ‘.‘;
while(!q.empty())
{
t = q.front(); q.pop();
for(i=0; i<8; ++i)
{
tx = t.x + dx[i];
ty = t.y + dy[i];
if(tx>=0&&tx<n&&ty>=0&&ty<m&&Map[tx][ty]==‘W‘)
{
Map[tx][ty] =‘.‘;
next.x = tx, next.y = ty;
q.push(next);
}
}
}
}
int main()
{
int i, j, total;
while(scanf("%d%d",&n,&m)!=EOF)
{
for(i=0; i<n; ++i) scanf("%s",Map[i]);
total = 0;
for(i=0; i<n; ++i)
{
for(j=0; j<m; ++j)
{
if(Map[i][j]==‘W‘)
{
total++;
bfs(i,j);
}
}
}
printf("%d\n",total);
}
return 0;
}
