Disjoint-set || [USACO18JAN] MooTube || BZOJ 5188 || Luogu P4185

Questions surface: [USACO18JAN] MooTube

answer:

On the side and ask all sorted and then each side to meet the current requirements are thrown in and check set for each inquiry and investigation to determine the current episode of nodes can be.

I'm bored to disjoint-set by adding a rank ordering, also opened O2, plus a quick read, and only 170ms, before too although in the first side, then goose was no way to row.

The above operation will do nothing

Code:

 1 #include<cstdio>
 2 #include<algorithm>
 3 using namespace std;
 4 inline int rd(){
 5     int x=0; char c=getchar();
 6     while(c<'0'||c>'9')c=getchar();
 7     while(c>='0'&&c<='9'){x=x*10+c-'0'; c=getchar();}
 8     return x;
 9 }
10 const int maxn=1e5+5,maxq=1e5+5;
11 int N,Q,a,b,c,fa[maxn],dep[maxn],f1,f2,now,cnt[maxn],ans[maxq];
12 struct Edge{ int from,to,dis; }edge[maxn];
13 struct Query_{ int k,x,id; }qry[maxq];
14 inline bool cmp1(const Edge&a,const Edge&b){ return a.dis>b.dis; }
15 inline bool cmp2(const Query_&a,const Query_&b){ return a.k>b.k; }
16 inline int getf(int a){
17     if(fa[a]==a) return a;
18     fa[a]=getf(fa[a]);
19     return fa[a];
20 }
21 int main(){
22     scanf("%d%d",&N,&Q);
23     for(int i=1;i<N;i++){
24         a=rd(); b=rd(); c=rd();
25         edge[i].from=a;
26         edge[i].to=b;
27         edge[i].dis=c;
28     }
29     for(int i=1;i<=Q;i++){
30         a=rd(); b=rd();
31         qry[i].k=a; qry[i].x=b; qry[i].id=i;
32     }
33     sort(edge+1,edge+N,cmp1);
34     sort(qry+1,qry+Q+1,cmp2);
35     for(int i=1;i<=N;i++) fa[i]=i,cnt[i]=1;
36     now=0;
37     for(int i=1;i<=Q;i++){
38         while(now+1<=N-1 && edge[now+1].dis>=qry[i].k){
39             now++;
40             f1=getf(edge[now].from);
41             f2=getf(edge[now].to);
42             if(f1!=f2){
43                 if(dep[f1]<dep[f2]){
44                     for [F1] = f2;
45                      cnt [f2] + = cnt [F1];
46                  }
 47                  else  if (dep [F1]> dep [f2]) {
 48                      makes [f2] = f1;
49                      cnt [F1] + = cnt [f2];
50                  }
 51                  else {
 52                      makes [F1] = f2;
53                      dep [f2] ++ ;
54                      cnt [f2] + = cnt [F1];
55                  }
 56              }
 57         }
58         ans[qry[i].id]=cnt[getf(qry[i].x)]-1;
59     }
60     for(int i=1;i<=Q;i++) printf("%d\n",ans[i]);
61     return 0;
62 }

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By:AlenaNuna