标签：char eterm can nal sep clu == new 建设
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
最小生成树裸题
#include<cmath> #include<cstdio> #include<cstring> #include<cstdlib> #include<iostream> #include<algorithm> #define N 510000 using namespace std; int n,m,x,y,s,fx,fy,xx[N],yy[N],fa[N]; int read() { int x=0,f=1; char ch=getchar(); while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1; ch=getchar();} while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘; ch=getchar();} return x*f; } struct Edge { int x,y; double z; }edge[N<<1]; int cmp(Edge a,Edge b) { return a.z<b.z; } int find(int x) { if(fa[x]==x) return x; fa[x]=find(fa[x]); return fa[x]; } int main() { n=read(),m=read(); for(int i=1;i<=n;i++) xx[i]=read(),yy[i]=read(); for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) if(i!=j) { ++s; edge[s].x=i; edge[s].y=j; edge[s].z=sqrt(pow(xx[i]-xx[j],2)+pow(yy[i]-yy[j],2)); } for(int i=1;i<=n;i++) fa[i]=i; double ans=0; for(int i=1;i<=m;i++) { x=read(),y=read(); fx=find(x),fy=find(y); fa[fx]=fy; } sort(edge+1,edge+1+s,cmp); for(int i=1;i<=s;i++) { x=edge[i].x,y=edge[i].y; fx=find(x),fy=find(y); if(fa[fx]==fy) continue; fa[fx]=fy; ans+=edge[i].z; } printf("%.2lf",ans); return 0; }
洛谷——P2872 [USACO07DEC]道路建设Building Roads
标签：char eterm can nal sep clu == new 建设
原文地址：http://www.cnblogs.com/z360/p/7468764.html
