水洼 POJ2386 挑战程序设计竞赛

1.题目原文

http://poj.org/problem?id=2386

Lake Counting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 30088		Accepted: 15038

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

Hint

OUTPUT DETAILS: 

There are three ponds: one in the upper left, one in the lower left,and one along the right side.
Source

USACO 2004 November

2.解题思路

采用深度优先搜索，从任意的w开始，不断把邻接的部分用'.'代替，1次DFS后与初始这个w连接的所有w就全都被替换成'.'，因此直到图中不再存在W为止，总共进行DFS的次数就是答案。8个方向对应8个状态转移，每个格子作为DFS的参数最多调用一次，因此时间复杂度为O(8nm)=O(nm)。

3.AC代码

#include <iostream>
#include<cstdio>
using namespace std;
#define maxn 105
char field[maxn][maxn];
int n,m;
void dfs(int x,int y)
{
field[x][y]='.';
//循环遍历八个方向
for(int dx=-1;dx<=1;dx++){
for(int dy=-1;dy<=1;dy++){
int nx=x+dx,ny=y+dy;
//判断(nx,ny)是否在园子里，以及是否有积水
if(0<=nx&&nx<n&&0<=ny&&ny<m&&field[nx][ny]=='W'){
dfs(nx,ny);
}
}
}
}
void solve()
{
int res=0;
for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
if(field[i][j]=='W'){
//从有积水的地方开始深搜
dfs(i,j);
res++;
}
}
}
printf("%d\n",res);
}
int main()
{
scanf("%d%d",&n,&m);
for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
cin>>field[i][j];
}
}
solve();
return 0;
}

4.类似题目：

http://poj.org/problem?id=1979

题目和上述解法类似，直接贴代码。

#include <iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define maxn 25
char field[maxn][maxn];
bool mark[maxn][maxn];
int n,m;
int res;
int dx[]={1,-1,0,0};
int dy[]={0,0,1,-1};
void dfs(int x,int y)
{
res++;
mark[x][y]=true;
//循环遍历八个方向
for(int i=0;i<4;i++){
int nx=x+dx[i],ny=y+dy[i];
if(0<=nx&&nx<n&&0<=ny&&ny<m&&field[nx][ny]=='.'&&!mark[nx][ny]){
//mark[x][y]=true;
dfs(nx,ny);
}
}
}
void solve()
{
int sx,sy;
memset(mark,0,sizeof(mark));
res=0;
for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
if(field[i][j]=='@'){
sx=i;
sy=j;
}
}
}
mark[sx][sy]=true;
dfs(sx,sy);
printf("%d\n",res);
}
int main()
{
while(cin>>m>>n){
if(n==0&&m==0) break;
for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
cin>>field[i][j];
}
}
solve();
}
return 0;
}