标签：ted with arm algo 4.0 logs i++ 格式 existing
Farmer John had just acquired several new farms! He wants to connect the farms with roads so that he can travel from any farm to any other farm via a sequence of roads; roads already connect some of the farms.
Each of the N (1 ≤ N ≤ 1,000) farms (conveniently numbered 1..N) is represented by a position (Xi, Yi) on the plane (0 ≤ Xi ≤ 1,000,000; 0 ≤ Yi ≤ 1,000,000). Given the preexisting M roads (1 ≤ M ≤ 1,000) as pairs of connected farms, help Farmer John determine the smallest length of additional roads he must build to connect all his farms.
给出nn个点的坐标,其中一些点已经连通,现在要把所有点连通,求修路的最小长度.
输入格式：
Line 1: Two space-separated integers: N and M
Lines 2..N+1: Two space-separated integers: Xi and Yi
输出格式：
4 1 1 1 3 1 2 3 4 3 1 4
4.00
一道kruskal裸题
注意精度问题，算两点间距离时要强制转换double类型
#include<iostream> #include<cstdio> #include<cmath> #include<algorithm> using namespace std; struct pos { int x,y; }a[1050]; double dis(pos a,pos b) { return sqrt((double)(a.x-b.x)*(double)(a.x-b.x)+(double)(a.y-b.y)*(double)(a.y-b.y)); } struct Edge { int u,v; double w; friend bool operator <(Edge a,Edge b){return a.w<b.w;} }e[1000050]; int father[1050]; int Find(int x) { if(father[x]==x) return x; return father[x]=Find(father[x]); } int main() { int n,m; scanf("%d%d",&n,&m); for(int i=1;i<=n;i++) scanf("%d%d",&a[i].x,&a[i].y); for(int i=1;i<=n;i++) father[i]=i; for(int i=1;i<=m;i++) { int x,y; scanf("%d%d",&x,&y); x=Find(x),y=Find(y); father[x]=y; } int tot=0; for(int i=1;i<=n;i++) for(int j=i;j<=n;j++) { tot++; e[tot].u=i; e[tot].v=j; e[tot].w=dis(a[i],a[j]); } sort(e+1,e+tot+1); double ans=0; int sum=0; for(int i=1;i<=tot;i++) { int u=Find(e[i].u),v=Find(e[i].v); if(u!=v) { father[u]=v; ans+=e[i].w; sum++; } } printf("%.2lf",ans); }
建造道路 (Building Roads, USACO 2007 Dec)
标签：ted with arm algo 4.0 logs i++ 格式 existing
原文地址：http://www.cnblogs.com/Position/p/7082248.html
