CF图论一-- Ice Skating--DFS||BFS||并查集

、、

C. Ice Skating

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Bajtek is learning to skate on ice. He's a beginner, so his only mode of transportation is pushing off from a snow drift to the north, east, south or west and sliding until he lands in another snow drift. He has noticed that in this way it's impossible to get from some snow drifts to some other by any sequence of moves. He now wants to heap up some additional snow drifts, so that he can get from any snow drift to any other one. He asked you to find the minimal number of snow drifts that need to be created.

We assume that Bajtek can only heap up snow drifts at integer coordinates.

Input

The first line of input contains a single integer n (1 ≤ n ≤ 100) — the number of snow drifts. Each of the following n lines contains two integers xi and yi (1 ≤ xi, yi ≤ 1000) — the coordinates of the i-th snow drift.

Note that the north direction coinсides with the direction of Oy axis, so the east direction coinсides with the direction of the Oxaxis. All snow drift's locations are distinct.

Output

Output the minimal number of snow drifts that need to be created in order for Bajtek to be able to reach any snow drift from any other one.

Examples

input

Copy

2
2 1
1 2

output

Copy

1

input

Copy

2
2 1
4 1

output

Copy

0

给出一些点的坐标，求出使得一个点至少存在一个和它平行的点，需要添加多少个点。

类似于BFS的油田块，求出可以当做整体的块数即可。

并查集做法

#include<bits/stdc++.h>
using namespace std;
#define maxn 100+66
int co[maxn];
vector<int>v[maxn];
set<int>s[maxn];
int n,m;
int vis[maxn];
struct node
{
    int x;
    int y;
}no[maxn];
int f[200];
int finds(int x)
{
    return f[x]==x?x:f[x]=finds(f[x]);
}
void unions(int x,int y)
{
    int fx=finds(x);
    int fy=finds(y);
    if(fx!=fy)
    {
        f[fx]=fy;
    }
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        f[i]=i;
        int a,b;
        scanf("%d%d",&no[i].x,&no[i].y);
    }
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=n;j++)
        {
            int x1=no[i].x;
            int y1=no[i].y;
            int x2=no[j].x;
            int y2=no[j].y;
            if(x1==x2||y1==y2)
            {
                unions(i,j);
            }
        }
    }
    for(int i=1;i<=n;i++)
    {
        finds(i);
    }
    int ans=0;
    for(int i=1;i<=n;i++)
    {
        if(f[i]==i)ans++;
    }
    printf("%d\n",ans-1);
}

DFS做法

#include<bits/stdc++.h>
using namespace std;
#define maxn 100+66
int n;
int x[maxn];
int y[maxn];
int vis[maxn];
int ans;
void dfs(int k)
{
    vis[k]=1;
    for(int i=1; i<=n; i++)
    {
        if(!vis[i]&&(x[i]==x[k]||y[i]==y[k]))
        {
            dfs(i);
        }
    }
}
int main()
{
    scanf("%d",&n);
    for(int i=1; i<=n; i++)
    {
        int a,b;
        scanf("%d%d",&x[i],&y[i]);
    }
    for(int i=1; i<=n; i++)
    {
        if(!vis[i])
        {
            ans++;//求块
            dfs(i);
        }
    }
    printf("%d\n",ans-1);
}

BFS做法

#include<bits/stdc++.h>
using namespace std;
#define maxn 100+66
int n;
int x[maxn];
int y[maxn];
int vis[maxn];
int ans;
void bfs(int k)
{
    queue<int>que;
    que.push(k);
    vis[k]=1;
    while(que.size())
    {
        int x1=que.front();
        que.pop();
        for(int i=1;i<=n;i++)
        {
            if(!vis[i]&&(x[i]==x[x1]||y[i]==y[x1]))
            {
                que.push(i);
                vis[i]=1;
            }
        }
    }
}
int main()
{
    scanf("%d",&n);
    for(int i=1; i<=n; i++)
    {
        int a,b;
        scanf("%d%d",&x[i],&y[i]);
    }
    for(int i=1; i<=n; i++)
    {
        if(!vis[i])
        {
            ans++;
            bfs(i);
        }
    }
    printf("%d\n",ans-1);
}