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Lake Counting

2024-07-12 05:57:14   222人阅读

Problem 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
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;
} 


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