有一个30000*N(i) 的列队,2种操作
1. M i,j i行移动到j行的末尾
2.C i,j 询问i行和j行的距离(如果在同一列)
#include <bits/stdc++.h> using namespace std ; const int N=3e4; int fa[N+2],L[N],n,sz[N]; int find(int x){ if(x==fa[x]) return x; int t=find(fa[x]); L[x]+=L[fa[x]]; return fa[x]=t; } signed main(){ int x,y,tes;char op; cin>>tes; for(int i=1;i<=N;i++) fa[i]=i,sz[i]=1; while(tes--){ cin>>op>>x>>y; int fx=find(x),fy=find(y); if(op=='M') L[fx]+=sz[fy],fa[fx]=fa[fy],sz[fy]+=sz[fx],sz[fx]=0; else{ if(fx==fy) cout<<abs(L[x]-L[y])-1<<endl; else cout<<-1<<endl; } } }
标签:sz,fx,int,fy,fa,银河,NOI2002,find,P1196 From: https://www.cnblogs.com/towboa/p/17215825.html