POJ 2349 Arctic Network (the bottom spanning tree + beg K large side)

Title Link: http://poj.org/problem?id=2349

The main topic: There are n outposts, and S satellite communication devices, any two satellite communication devices can communicate through satellites, regardless of their location. Otherwise, only the distance between two outposts does not exceed D to communicate over the radio. Find the smallest d that enables all outposts to communicate directly or indirectly.

Problem-Solving ideas: The topic requires that all outposts can communicate directly or indirectly, then the equivalent of n points connected, at least the need to n-1 edge. N points can be divided into S groups, each intra-group unlimited communication, between the regiment and the group through satellite communications. Then it is equivalent to use the S satellite device to establish S-1 edge, with radio communication to establish n-1-(s-1) ==n-s edge, because the satellite communication is no distance limit that can choose the largest s-1 edge, that D is the left-n-S bar edge of the longest side distance.

Code:

1#include <iostream>2#include <cstdio>3#include <math.h>4#include <cstring>5#include <algorithm>6 using namespacestd;7 Const intn=1e3+5;8 9 structnode2{Ten     intx, y; One }a[n]; A  - structnode{ -     intx, y; the     Doubledis; - node () {} -NodeintXintYDoubledis) { -          This->x=x; +          This->y=y; -          This->dis=dis; +     } A     BOOL operator< (ConstNode &b)Const{ at         returndis<B.dis; -     } -}edge[n*N]; - intRoot[n]; -  - intFindintx) { in     returnroot[x]==x?x:root[x]=find (Root[x]); - } to  + intMain () { -     intT; thescanf"%d",&t); *      while(t--){ $         intS,n;Panax Notoginsengscanf"%d%d",&s,&n); -          for(intI=1; i<=n;i++){ thescanf"%d%d",&a[i].x,&a[i].y); +root[i]=i; A         } the         intCnt=0; +          for(intI=1; i<=n;i++){ -              for(intj=i+1; j<=n;j++){ $                 DoubleDIS=SQRT (1.0* (a[i].x-a[j].x) * (a[i].x-a[j].x) + (A[I].Y-A[J].Y) * (a[i].y-a[j].y)); $edge[++cnt]=node (i,j,dis); -             } -         } theSort (edge+1, edge+1+CNT); -         Wuyi         intedge_num=0; the         Doubleans; -          for(intI=1; i<=cnt;i++){ Wu             intx=find (edge[i].x); -             inty=find (EDGE[I].Y); About             if(x!=y) { $edge_num++; -root[x]=y; -                 //Remove the maximum s-1 edge from the N-1 bar in the smallest spanning tree (because there is a S satellite station, which is equivalent to the S-point, then the s-1 edge) -                 //in the remaining n-s bars, the largest side length is the one that is asked A                 if(edge_num==n-s) { +ans=Edge[i].dis; the                      Break; -                 } $             } the         } theprintf"%.2f\n", ans); the     } the     return 0; -}

POJ 2349 Arctic Network (the bottom spanning tree + beg K large side)