【POJ2386】Lake Counting(深搜与广搜)

Lake Counting

Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 17917 Accepted: 9069

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

这一题就是油田问题的翻版，题目大意就是说找出W组成的块（一个块是几个W连接在一起的，可能通过斜线也能连接。）有几个。
题目很简单，循环找到W然后将与其连接的置为‘.’。块数+1，就可以了。
下面是两种做法：
深搜：

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
#include "math.h"

using namespace std;

const int maxn = 1e2+5;

char map[maxn][maxn];
int n,m;
int changeX[8]={
   
   0,0,1,1,1,-1,-1,-1};
int changeY[8]={
   
   1,-1,1,-1,0,1,-1,0};

int check(int x,int y){
    if(x<0||y<0||x>n||y>m||map[x][y]!='W') return 0;
    return 1;
}

void dfs(int x,int y){
    int i;
    for(i=0 ; i<8 ; i++){
        int nowX = x+changeX[i];
        int nowY = y+changeY[i];
        if(check(nowX,nowY)){
            map[nowX][nowY]='.';
            dfs(nowX,nowY);
        }
    }
}

int main(){
    while(~scanf("%d%d",&n,&m)){
        int ans=0;
        for(int i=0 ; i<n ; i++){
            for(int j=0 ; j<m ; j++){
                cin>>map[i][j];
            }
        }
        for(int i=0 ;i<n ; i++){
            for(int j=0 ;j<m ; j++){
                if(map[i][j]=='W'){
                    dfs(i,j);
                    ans++;
                }
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}

广搜：

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
#include "math.h"
#include "queue"

using namespace std;

const int maxn = 1e2+5;

char map[maxn][maxn];
int n,m;
int ans;
int changeX[8]={
   
   0,0,1,1,1,-1,-1,-1};
int changeY[8]={
   
   1,-1,1,-1,0,1,-1,0};

struct FYJ{
    int x;
    int y;
};

int check(int x,int y){
    if(x<0||y<0||x>n||y>m||map[x][y]!='W')
    return 0;
    return 1;
}

void BFS(int x,int y){
    queue<FYJ>q;
    FYJ now,to;
    now.x=x; now.y=y;
    q.push(now);
    while(!q.empty()){
        now = q.front();
        q.pop();
        for(int i=0 ; i<8 ; i++){
            to.x=now.x+changeX[i];
            to.y=now.y+changeY[i];
            if(check(to.x,to.y)){
                q.push(to);
                map[to.x][to.y]='.';
            }
        }
    }
}

int main(){
    while(~scanf("%d%d",&n,&m)){
        ans = 0;
        for(int i=0 ; i<n ; i++){
            for(int j=0 ; j<m ; j++){
                cin>>map[i][j];
            }
        }
        for(int i=0 ; i<n ; i++){
            for(int j=0 ; j<m ; j++){
                if(map[i][j]=='W'){
                    ans++;
                    BFS(i,j);
                }
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}