POJ-2349 Arctic Network 最小生成树

来源：互联网发布：js decode 编辑：程序博客网时间：2024/05/29 14:34

比较简单一题，求出最小生成树后，第s条长的线段即可，问题是我特么WA了N次不知道原因，最后找来正确代码单步调试差错，最后发现是因为POJ只能用%f输出！！！！老子居然该死的把这一点忘了，还他妈一直想不出来，真是醉了。

#include <iostream>#include <cstring>#include <cmath>#include <cstdio>#include <queue>#include <algorithm>#include <map>#include <vector>using namespace std;const int maxn=505; struct Point{int x;int y;}point[maxn];struct Edge{int from;int to;double cost;int visit;}edge[maxn*maxn];int m;int father[maxn];double dis(Point a,Point b){return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}bool cmp(Edge a,Edge b){return a.cost<b.cost;}int find(int x){if(father[x]==x) return x;     father[x]=find(father[x]);     return father[x];}bool Union(int x,int y){int fx=find(x);int fy=find(y);if(fx==fy){return false;}else{father[fx]=fy;}return true;}double kruskal( int n ) {     int  i , j = 0 ;     double sum = 0 ;     for( i = 0 ; i < n ; i ++ )         father[i] = i ;     sort( edge , edge +m , cmp ) ;     for( i = 0 ; i < m && j < n ; i ++ )     {         if( Union( edge[i].from , edge[i].to ) )         {             sum += edge[i].cost  ;             edge[i].visit  = 1 ;             j ++ ;         }     }     return sum ; } int main(){int t;int s,p;int i,j;scanf("%d",&t);while(t--){m=0;scanf("%d%d",&s,&p);for(i=0;i<p;i++){scanf("%d%d",&point[i].x,&point[i].y);}for(i=0;i<p-1;i++){for(j=i+1;j<p;j++){edge[m].from=i;edge[m].to=j;edge[m].cost=dis(point[i],point[j]);edge[m].visit=0;m++;}}kruskal(p);int u;for(i=m-1;i>=0;i--){             if( edge[i].visit )             {                 s -- ;                 if( s == 0 ) break ;             }         }         printf( "%.2f\n" ,edge[i].cost ) ; }return 0;}

0 0