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: 65536KTotal 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 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.

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 105char 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 25char 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;}