编程爱好者联盟 2017-03-04
用并查集记录每个联通块的根节点,每个联通块建一棵线段树,合并时合并线段树就可以了。
代码:
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 using namespace std;
5 #define N 100010
6 struct node{
7 int l,r,x;
8 }c[N*20];
9 int i,j,k,n,m,l,r,a[N],x,y,Rt[N],Num,b[N],f[N],f1,f2;
10 char s[2];
11 inline int Find(int x){return x==f[x]?x:f[x]=Find(f[x]);}
12 inline void Update(int& Node,int l,int r,int x){
13 if(!Node)Node=++Num;
14 if(l==r){c[Node].x++;return;}
15 int Mid=l+r>>1;
16 if(Mid<x)Update(c[Node].r,Mid+1,r,x);else Update(c[Node].l,l,Mid,x);
17 c[Node].x=c[c[Node].l].x+c[c[Node].r].x;
18 }
19 inline int Merge(int x,int y){
20 if(!y)return x;
21 if(!x)return y;
22 c[x].l=Merge(c[x].l,c[y].l);
23 c[x].r=Merge(c[x].r,c[y].r);
24 c[x].x=c[c[x].l].x+c[c[x].r].x;
25 return x;
26 }
27 inline int Query(int Node,int l,int r,int R){
28 if(l>R||!Node)return 0;
29 if(r<=R)return c[Node].x;
30 int Mid=l+r>>1;
31 return Query(c[Node].l,l,Mid,R)+Query(c[Node].r,Mid+1,r,R);
32 }
33 inline int Get_Ans(int x,int k){
34 int l=1,r=n,Mid;
35 while(l<=r){
36 Mid=l+r>>1;
37 if(Query(Rt[x],1,n,Mid)>=k)r=Mid-1;else l=Mid+1;
38 }
39 if(l>n)return -1;return b[l];
40 }
41 int main()
42 {
43 scanf("%d%d",&n,&m);
44 for(i=1;i<=n;i++)scanf("%d",&a[i]),b[a[i]]=i,f[i]=i;
45 while(m--)scanf("%d%d",&x,&y),f[Find(x)]=Find(y);
46 for(i=1;i<=n;i++)Update(Rt[Find(i)],1,n,a[i]);
47 scanf("%d",&m);
48 while(m--){
49 scanf("%s%d%d",s,&x,&y);
50 if(s[0]=='Q')printf("%d\n",Get_Ans(Find(x),y));else{
51 f1=Find(x);f2=Find(y);
52 if(f1!=f2){
53 f[f1]=f2;
54 Rt[f2]=Merge(Rt[f1],Rt[f2]);
55 }
56 }
57 }
58 return 0;
59 } bzoj2733