poj2386——油田问题（简单搜索）

Lake Counting

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 35186 Accepted: 17477

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

这是一道很水的题

复习一下搜索，经典的油田问题，用的也是最简单的递归法。。。也要尝试一下非递归DFS和BFS的

//#include<bits/stdc++.h>//poj不认。。。#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<algorithm>#include<stack>#include<vector>#include<map>#include<set>const int M=1005;using namespace std;char f[M][M];bool vis[M][M];int n,m;bool judge(int i,int j){    if(i<=0||j<=0||i>n||j>m||f[i][j]=='.')        return false;    else        return true;}void dfs(int x,int y){    f[x][y]='.';    for(int dx=-1;dx<=1;dx++){        for(int dy=-1;dy<=1;dy++){            int x1=x+dx;            int y1=y+dy;            if(0<=x1&&x1<n&&0<=y1&&y1<m&&f[x1][y1]=='W')                dfs(x1,y1);        }    }    return;}int main(){    freopen("F://1.txt","r",stdin);    int i,j,k;    scanf("%d %d",&n,&m);///个人习惯了scanf，cin流输入可以避免回车问题        getchar();//这里要吸收一个回车。。。        for(i=0;i<n;i++){            for(j=0;j<m;j++){                scanf("%c",&f[i][j]);            }            getchar();//还有这里。。。        }        int ans=0;        for(i=0;i<n;i++){            for(j=0;j<m;j++){                if(f[i][j]=='W'){                    dfs(i,j);                    ans++;                }            }        }        printf("%d\n",ans);    return 0;}/*10 12W........WW..WWW.....WWW....WW...WW..........WW..........W....W......W...W.W.....WW.W.W.W.....W..W.W......W...W.......W.*/

栈实现（有bug，样例结果为5，目前无解。。。）

#include<iostream>#include<cstdio>#include<cstring>#include<string>#include <algorithm>#include <stack>#include <queue>#include <vector>#include <map>#include <set>using namespace std;const int M=1005;char f[M][M];bool vis[M][M];int n,m;struct point{    int x,y;};stack <point> s;int d[8][2] = {{0,1},{0,-1},{1,0},{-1,0},{1,1},{1,-1},{-1,1},{-1,-1}};bool judge(int i,int j){    if(i<=0||j<=0||i>n||j>m||f[i][j]=='.')        return false;    else        return true;}void dfs(int x,int y){    /*f[x][y]='.';    for(int dx=-1;dx<=1;dx++){        for(int dy=-1;dy<=1;dy++){            int x1=x+dx;            int y1=y+dy;            if(0<=x1&&x1<n&&0<=y1&&y1<m&&f[x1][y1]=='W')                dfs(x1,y1);        }    }    return;*/    int i,j,k;    point t;    t.x=x;    t.y=y;    f[t.x][t.y]='.';    s.push(t);    while(!s.empty()){        point temp=s.top();        f[temp.x][temp.y]='.';        s.pop();        for(i=0;i<8;i++){            point t1;            t1.x=temp.x+d[i][0];            t1.y=temp.y+=d[i][1];            if(t1.x>=0&&t1.x<n&&t1.y>=0&&t1.y<m&&f[t1.x][t1.y]=='W'){                f[t1.x][t1.y]='.';                s.push(t1);            }            /*else{                s.pop();            }*/        }    }    return;}int main(){    freopen("F://1.txt","r",stdin);    int i,j,k;    //scanf("%d %d",&n,&m);    while(!s.empty()){        s.pop();    }    cin>>n>>m;        //getchar();        for(i=0;i<n;i++){            for(j=0;j<m;j++){               // scanf("%c",&f[i][j]);               cin>>f[i][j];            }           // getchar();        }        int ans=0;        for(i=0;i<n;i++){            for(j=0;j<m;j++){                if(f[i][j]=='W'){                    dfs(i,j);                    ans++;                }            }        }        printf("%d\n",ans);    return 0;}

BFS:

#include<iostream>#include<cstdio>#include<cstring>#include<string>#include <algorithm>#include <stack>#include <queue>#include <vector>#include <map>#include <set>using namespace std;const int M=1005;char f[M][M];bool vis[M][M];int n,m;struct point{    int x,y;};queue <point> q;int d[8][2] = {{0,1},{0,-1},{1,0},{-1,0},{1,1},{1,-1},{-1,1},{-1,-1}};bool judge(int i,int j){    if(i<0||j<0||i>n-1||j>m-1||f[i][j]=='.')        return false;    else        return true;}void bfs(int x,int y){    point p;    p.x=x;p.y=y;    q.push(p);    f[x][y]='.';    //更改遍历过的位置的状态    int i,j,k;    while(!q.empty()){        point t1;        t1=q.front();        q.pop();        for(i=0;i<8;i++){            point t2;            t2.x=t1.x+d[i][0];            t2.y=t1.y+d[i][1];            if(judge(t2.x,t2.y)){                q.push(t2);                f[t2.x][t2.y]='.';//更改遍历过的点的状态            }        }    }    return;}int main(){    freopen("F://1.txt","r",stdin);    int i,j,k;    while(!q.empty()){        q.pop();    }    //scanf("%d %d",&n,&m);    cin>>n>>m;        //getchar();        for(i=0;i<n;i++){            for(j=0;j<m;j++){               // scanf("%c",&f[i][j]);               cin>>f[i][j];            }           // getchar();        }        int ans=0;        for(i=0;i<n;i++){            for(j=0;j<m;j++){                if(f[i][j]=='W'){                    bfs(i,j);                    ans++;                }            }        }        printf("%d\n",ans);    return 0;}