AtCoder Regular Contest 076 D

D - Built?

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Time limit : 2sec / Memory limit : 256MB

Score : 500 points

Problem Statement

There are N towns on a plane. The i-th town is located at the coordinates (xi,yi). There may be more than one town at the same coordinates.

You can build a road between two towns at coordinates (a,b) and (c,d) for a cost of min(|a−c|,|b−d|) yen (the currency of Japan). It is not possible to build other types of roads.

Your objective is to build roads so that it will be possible to travel between every pair of towns by traversing roads. At least how much money is necessary to achieve this?

Constraints

• 2≤N≤105
• 0≤xi,yi≤109
• All input values are integers.

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Input

Input is given from Standard Input in the following format:

N
x1 y1
x2 y2
:
xN yN

Output

Print the minimum necessary amount of money in order to build roads so that it will be possible to travel between every pair of towns by traversing roads.

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Sample Input 1

3
1 5
3 9
7 8

Sample Output 1

3

Build a road between Towns 1 and 2, and another between Towns 2 and 3. The total cost is 2+1=3 yen.

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Sample Input 2

6
8 3
4 9
12 19
18 1
13 5
7 6

Sample Output 2

8

分别按照x和y坐标排下序，相邻两点建边，最后最小生成树的边集一定是这些边里面的。
其实参照prim算法求解的过程就明白啦，最后我是用Kruskal求解的。

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<algorithm>
#include<iostream>
#include<queue>
#include<map>
#include<cmath>
#include<set>
#include<stack>
#define ll long long
#define pb push_back
#define max(x,y) ((x)>(y)?(x):(y))
#define min(x,y) ((x)>(y)?(y):(x))
#define cls(name,x) memset(name,x,sizeof(name))
using namespace std;
const int inf=1e9+10;
const int maxn=1e5+10;
const int maxm=20;
const int mod=1e9+7;
const double pi=acos(-1.0);
int n;
struct node
{
    int x,y;
    int id;
}town[maxn];
struct edge
{
    int st,ed;
    int cost;
}e[maxn*4];
int fa[maxn];
bool cmp1(const node &a,const node &b)
{
    return a.x<b.x;
}
bool cmp2(const node &a,const node &b)
{
    return a.y<b.y;
}
bool cmp3(const edge &a,const edge &b)
{
    return a.cost<b.cost;
}
int Find(int x)
{
    if(fa[x]!=x)
        fa[x]=Find(fa[x]);
    return fa[x];
}
int main()
{
    //freopen("in.txt","r",stdin);
    while(~scanf("%d",&n))
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%d %d",&town[i].x,&town[i].y);
            town[i].id=i;
        }
        int k=0;
        sort(town+1,town+1+n,cmp1);
        for(int i=1;i<=n-1;i++)
        {
            e[k].st=town[i].id;
            e[k].ed=town[i+1].id;
            e[k++].cost=abs(town[i].x-town[i+1].x);
        }
        sort(town+1,town+1+n,cmp2);
        for(int i=1;i<=n-1;i++)
        {
            e[k].st=town[i].id;
            e[k].ed=town[i+1].id;
            e[k++].cost=abs(town[i].y-town[i+1].y);
        }
        sort(e,e+k,cmp3);
        for(int i=1;i<=n;i++)
            fa[i]=i;
        int ans=0;
        for(int i=0;i<k;i++)
        {
            int a=Find(e[i].st),b=Find(e[i].ed);
            if(a!=b)
            {
                ans+=e[i].cost;
                fa[b]=a;
            }
        }
        printf("%d
",ans);
    }
    return 0;
}