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最新推荐文章于 2022-07-24 19:58:51 发布

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解决一个经典的图遍历问题——如何计算给定矩阵中相连的‘W’（代表水）形成的独立池塘数量。该文提供了两种不同的实现方法，帮助读者理解深度优先搜索算法。

Lake Counting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 35731		Accepted: 17734

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

题意：有一个大小为n*m的园子，求出水洼的个数，一个W是一个水洼，多个连着的W也是一个水洼，也就是一个或者多个连着的W组成一个水洼，W表示积水，'.'表示没有积水

这道题目贼坑！算是深搜里的水题了吧！

我用了两种表示方式做

（1）

#include<stdio.h>
int n,m;
char field[1000][1000];
int res;
void dfs(int x,int y )
{
    int dx,dy,nx,ny;
    field[x][y]='.';//讲现在所在的位置替换为'.'
    for(dx=-1;dx<=1;dx++)
    {
        for(dy=-1;dy<=1;dy++)
        {
            nx=dx+x;
            ny=dy+y;
            if(0<=nx&&nx<n&&0<=ny&&ny<m&&field[nx][ny]=='W')dfs(nx,ny);//判断（nx,ny)是不是在园子里，是否有积水
        }
    }
}
int main()
{
    int i,j;
    scanf("%d%d",&n,&m);

        for(i=0;i<n;i++)
        {
            getchar();
            for(j=0;j<m;j++)
            {
                scanf("%c",&field[i][j]);
            }
        }
        res=0;
        for(i=0;i<n;i++)
        {
            for(j=0;j<m;j++)
            {
                if(field[i][j]=='W')
                {
                    dfs(i,j);
                    res++;
                }
            }
        }
        printf("%d\n",res);
    return 0;
}

（2）

#include<stdio.h>
char map[1000][1000];
int n,m;
int a[8][2]={{1,1},{1,0},{1,-1},{0,1},{0,-1},{-1,0},{-1,1},{-1,-1}};
int ans;
void dfs(int x,int y)
{
    int i,dx,dy;
    map[x][y]='.';
    for(i=0;i<8;i++)
    {
        dx=x+a[i][0];
        dy=y+a[i][1];
        if(dx>=0&&dx<n&&dy>=0&&dy<m&&map[dx][dy]=='W') dfs(dx,dy);
    }
}
int main()
{
    int i,j;
    scanf("%d%d",&n,&m);
        for(i=0;i<n;i++)
        {
            getchar();
            for(j=0;j<m;j++)
            {
                scanf("%c",&map[i][j]);
            }
        }
        ans=0;
        for(i=0;i<n;i++)
        {
            for(j=0;j<m;j++)
            {
                if(map[i][j]=='W')
                {
                    dfs(i,j);
                    ans++;
                }
            }
        }
        printf("%d\n",ans);
    return 0;
}