性质：最小生成树上任意两点间的最大边权，一定是这两点间所有路径的最大边权中最小的。证明显然。
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #define rep(i,l,r) for (int i=(l); i<=(r); i++) 5 using namespace std; 6 7 const int N=1010,M=1000010; 8 int n,m,tot,ans,mx,k,a[N],fa[N]; 9 struct P{ int x,y; }p[N]; 10 struct E{ int u,v,w; }e[M]; 11 bool operator <(const E &a,const E &b){ return a.w<b.w; } 12 int sqr(int x){ return x*x; } 13 int get(int x){ return (fa[x]==x) ? x : fa[x]=get(fa[x]); } 14 15 int main(){ 16 scanf("%d",&m); 17 rep(i,1,m) scanf("%d",&a[i]); 18 scanf("%d",&n); 19 rep(i,1,n) scanf("%d%d",&p[i].x,&p[i].y),fa[i]=i; 20 rep(i,1,n-1) rep(j,i+1,n) e[++tot]=(E){i,j,sqr(p[i].x-p[j].x)+sqr(p[i].y-p[j].y)}; 21 sort(e+1,e+tot+1); 22 rep(i,1,tot){ 23 int u=e[i].u,v=e[i].v; 24 if (get(u)==get(v)) continue; 25 k++; mx=max(mx,e[i].w); 26 if (k==n-1) break; 27 fa[get(u)]=get(v); 28 } 29 rep(i,1,m) if (sqr(a[i])>=mx) ans++; 30 printf("%d ",ans); 31 return 0; 32 }