題目描述

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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輸入

Input is given from Standard Input in the following format:

N
x1 y1
x2 y2
:
xN yN

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輸出

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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樣例輸入

3 1 5 3 9 7 8

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樣例輸出

3

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提示

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

今日大兇......總之審題，數據范圍都弄不清了

這題初看感覺是最短路，但是每個點連起來就是1e10了

題意是把每個點連起來，需要花費多少，花費就是min(|a?c|,|b?d|)

所以并不需要計算每個點的距離，只需x和y分別排序一下就好，最短距離只有可能從按大小排序的這些點的距離選

然后是Kruskal算法?https://blog.csdn.net/liangzhaoyang1/article/details/51169090

（1） 將全部邊按照權值由小到大排序。
（2） 按順序（邊權由小到大的順序）考慮每條邊，只要這條邊和我們已經選擇的邊不構成圈，就保留這條邊，否則放棄這條邊。

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#include <bits/stdc++.h> using namespace std; const int mod=1e9+7; const int maxn=1e5+7; int book[maxn],n; // void init() {for(int i=1;i<=n;i++) book[i]=i; } int getfa(int x){if(book[x]!=x) book[x]=getfa(book[x]);return book[x]; } void mergexy(int x,int y) {book[y]=x; } // struct flv {int u,v,z; }f[maxn*2]; struct node {int x,y,num; }s[maxn]; bool cmp1(node a,node b) {return a.x<b.x; } bool cmp2(node a,node b) {return a.y<b.y; } bool cmp3(flv a,flv b) {return a.z<b.z; } int main(){scanf("%d",&n);for(int i=1;i<=n;i++){scanf("%d %d",&s[i].x,&s[i].y);s[i].num=i;}sort(s+1,s+n+1,cmp1);int cnt=0;for(int i=1;i<n;i++){f[++cnt].u=s[i].num;f[cnt].v=s[i+1].num;f[cnt].z=min(s[i+1].x-s[i].x,abs(s[i+1].y-s[i].y));}sort(s+1,s+n+1,cmp2);for(int i=1;i<n;i++){f[++cnt].u=s[i].num;f[cnt].v=s[i+1].num;f[cnt].z=min(s[i+1].y-s[i].y,abs(s[i+1].x-s[i].x));}sort(f+1,f+cnt+1,cmp3);init();int number=0;int sum=0;for(int i=1;i<=cnt;i++){int p=getfa(f[i].u),q=getfa(f[i].v);if(p!=q){mergexy(p,q);number++;sum+=f[i].z;}if(number==n) break;}printf("%d\n",sum);return 0; } View Code

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轉載于:https://www.cnblogs.com/smallocean/p/9271500.html

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