uva1479 - Graph and Queries Treap名次树

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原创最新推荐文章于 2026-06-08 12:06:32 发布 564 阅读 AI 写同款· GEO 优化 ›

Bridge - 判断图是否连通 - 图的遍历 or 并查集搜索 or 强连通算法 You are given an undirected connected graph with N vertices and M edges that does not contain self-loops and double edges. The i-th edge (1≤i≤M) connects Vertex ai and Vertex bi. An edge whose removal disconnec 阅读详情

You are given an undirected graph with N vertexes and M edges. Every vertex in this graph has an integer value assigned to it at the beginning. You're also given a sequence of operations and you need to process them as requested. Here's a list of the possible operations that you might encounter:

Deletes an edge from the graph.
The format is [D X], where X is an integer from 1 to M, indicating the ID of the edge that you should delete. It is guaranteed that no edge will be deleted more than once.
Queries the weight of the vertex with K-th maximum value among all vertexes currently connected with vertex X (including X itself).
The format is [Q X K], where X is an integer from 1 to N, indicating the id of the vertex, and you may assume that K will always fit into a 32-bit signed integer. In case K is illegal, the value for that query will be considered as undefined, and you should return 0 as the answer to that query.
Changes the weight of a vertex.
The format is [C X V], where X is an integer from 1 to N, and V is an integer within the range [-10⁶, 10⁶].

The operations end with one single character, E, which indicates that the current case has ended. For simplicity, you only need to output one real number - the average answer of all queries.

Input

There are multiple test cases in the input file. Each case starts with two integers N and M (1 $\le$ N $\le$ 2 * 10⁴, 0 $\le$ M $\le$ 6 * 10⁴), the number of vertexes in the graph. The next N lines describes the initial weight of each vertex (- 10⁶ $\le$ [weight][i] $\le$ 10⁶). The next part of each test case describes the edges in the graph at the beginning. Vertexes are numbered from 1 to N. The last part of each test case describes the operations to be performed on the graph. It is guaranteed that the number of query operations [Q X K] in each case will be in the range [1, 2 * 10⁵], and there will be no more than 2 * 10⁵ operations that change the values of the vertexes [C X V].

There will be a blank line between two successive cases. A case with N = 0, M = 0 indicates the end of the input file and this case should not be processed by your program.

Output

For each test case, output one real number - the average answer of all queries, in the format as indicated in the sample output. Please note that the result is rounded to six decimal places.

Explanation for samples:

For the first sample:

D 3 - deletes the 3rd edge in the graph (the remaining edges are (1, 2) and (2, 3))

Q 1 2 - finds the vertex with the second largest value among all vertexes connected with 1. The answer is 20.

Q 2 1 - finds the vertex with the largest value among all vertexes connected with 2. The answer is 30.

D 2 - deletes the 2nd edge in the graph (the only edge left after this operation is (1, 2))

Q 3 2 - finds the vertex with the second largest value among all vertexes connected with 3. The answer is 0 (Undefined).

C 1 50 - changes the value of vertex 1 to 50.

Q 1 1 - finds the vertex with the largest value among all vertex connected with 1. The answer is 50.

E - This is the end of the current test case. Four queries have been evaluated, and the answer to this case is (20 + 30 + 0 + 50) / 4 = 25.000.

For the second sample, caution about the vertex with same weight:

Q 1 1 - the answer is 20

Q 1 2 - the answer is 20

Q 1 3 - the answer is 10

Sample Input

Sample Output

Case 1: 25.000000 
Case 2: 16.666667

N个节点M条边的无向图，每个节点都有一个整数权值，有三种操作。

D X 删除ID为X的边。Q X K，计算与节点X连通的节点中（包括本身）第K大的值。如果不存在为0。C X V，把结点X的权值改为V。输出所有Q指令计算结果的平均值。

Treap名次树就是排序二叉树多了个rank，这棵树的rank值符合堆的性质，父亲的rank大。这是为了避免二叉树插入顺序的极端情况，每新建结点的时候随机一个rank值，和把所有结点随机排序再插入等价。

插入的时候先按一般排序二叉树插入，再判断根结点和插入这个结点的子树的根结点的rank，如果不符合堆的性质，就旋转。旋转分为左旋和右旋，左旋是把根变成右结点的左结点，原来右结点的左结点变成原来根的右结点，原来根的右结点变成根。右旋是把根变成左结点的右结点，原来左结点的右结点变成原来根的左结点，原来根的左结点变成根。

离线处理，把所有操作反过来。最开始把最后剩下的边全加到图里。遇到D操作就相当于把这条边加到图里，遇到C操作就是把V改成X。这样倒着来是因为删边不好删，加边只需要找出两个端点所在的集合合并就行了。注意把结点数少的合并到结点数多的集合中，对于任意结点来说，每次移动到新的树汇中时，该结点所在的树的大小至少加倍，所以任意结点至多被移动logN次，每次移动需要O(logN)的时间，因此总时间复杂度为O（nlog^2n）。

#include<iostream>
#include<algorithm>
#include<queue>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<set>
#include<map>
#include<stack>
#define INF 0x3f3f3f3f
#define MOD 1000000007
#define MAXNODE 8*MAXN
#define eps 1e-10
using namespace std;
typedef long long LL;

const int MAXN=20010;
const int MAXM=60010;
const int MAXC=500010;

int N,M,K,query_cnt;
LL query_tot;
int pa[MAXN],weight[MAXN],removed[MAXM];
char str[10];

struct Node{
    Node* ch[2];    //左右子树
    int r;          //优先级
    int v;          //值
    int s;          //结点总数

    Node(int v):v(v){
        ch[0]=ch[1]=NULL;
        r=rand();   //随机一个r值
        s=1;
    }
    bool operator < (const Node& rhs) const{
        return r<rhs.r;
    }
    int cmp(int x) const{
        if(x==v) return -1;
        return x<v?0:1;
    }
    void maintain(){
        s=1;
        if(ch[0]!=NULL) s+=ch[0]->s;
        if(ch[1]!=NULL) s+=ch[1]->s;
    }
}* root[MAXN];

struct Edge{
    int from,to;
}edge[MAXM];

struct Command{
    char op;
    int x,k;
}command[MAXC];

//d=0左旋,d=1右旋
void rotate(Node* &o,int d){
    Node* k=o->ch[d^1];
    o->ch[d^1]=k->ch[d];
    k->ch[d]=o;
    o->maintain();
    k->maintain();
    o=k;
}

void insert(Node* &o,int x){
    if(o==NULL) o=new Node(x);
    else{
        int d=(x<o->v?0:1);
        insert(o->ch[d],x);
        if(o->ch[d]->r>o->r) rotate(o,d^1);
    }
    o->maintain();
}

void remove(Node* &o,int x){
    int d=o->cmp(x);
    if(d==-1){
        Node* u=o;
        if(o->ch[0]!=NULL&&o->ch[1]!=NULL){
            int d2=(o->ch[0]->r>o->ch[1]->r?1:0);
            rotate(o,d2);
            remove(o->ch[d2],x);
        }
        else{
            if(o->ch[0]==NULL) o=o->ch[1];
            else o=o->ch[0];
            delete u;
        }
    }
    else remove(o->ch[d],x);
    if(o!=NULL) o->maintain();
}

int findset(int x){
    return x==pa[x]?x:pa[x]=findset(pa[x]);
}

//把src的结点移动到dest
void mergeto(Node* &src,Node* &dest){
    if(src->ch[0]!=NULL) mergeto(src->ch[0],dest);
    if(src->ch[1]!=NULL) mergeto(src->ch[1],dest);
    insert(dest,src->v);
    delete src;
    src=NULL;
}

int kth(Node* o,int k){
    if(o==NULL||k<=0||k>o->s) return 0;
    int s=(o->ch[1]==NULL?0:o->ch[1]->s);
    if(k==s+1) return o->v;
    else if(k<=s) return kth(o->ch[1],k);
    else return kth(o->ch[0],k-s-1);
}

void add_edge(int x){
    int u=findset(edge[x].from),v=findset(edge[x].to);
    if(u!=v){
        if(root[u]->s<root[v]->s){
            pa[u]=v;
            mergeto(root[u],root[v]);
        }
        else{
            pa[v]=u;
            mergeto(root[v],root[u]);
        }
    }
}

void query(int x,int k){
    query_cnt++;
    query_tot+=kth(root[findset(x)],k);
}

void change_weight(int x,int v){
    int u=findset(x);
    remove(root[u],weight[x]);
    insert(root[u],v);
    weight[x]=v;
}

void removetree(Node* x){
    if(x->ch[0]!=NULL) removetree(x->ch[0]);
    if(x->ch[1]!=NULL) removetree(x->ch[1]);
    delete x;
    x=NULL;
}

int main(){
    freopen("in.txt","r",stdin);
    int cas=0;
    while(scanf("%d%d",&N,&M)!=EOF&&(N||M)){
        for(int i=1;i<=N;i++) scanf("%d",&weight[i]);
        for(int i=1;i<=M;i++) scanf("%d%d",&edge[i].from,&edge[i].to);
        memset(removed,0,sizeof(removed));
        K=0;
        int x,k,v;
        while(scanf("%s",str),str[0]!='E'){
            k=v=0;
            scanf("%d",&x);
            if(str[0]=='D') removed[x]=1;
            else if(str[0]=='Q') scanf("%d",&k);
            //k->v
            else if(str[0]=='C'){
                scanf("%d",&v);
                k=weight[x];
                weight[x]=v;
            }
            command[K++]=(Command){str[0],x,k};
        }
        for(int i=1;i<=N;i++){
            pa[i]=i;
            if(root[i]!=NULL) removetree(root[i]);
            root[i]=new Node(weight[i]);
        }
        for(int i=1;i<=M;i++) if(!removed[i]){
            add_edge(i);
        }
        query_tot=query_cnt=0;
        for(int i=K-1;i>=0;i--){
            if(command[i].op=='D') add_edge(command[i].x);
            else if(command[i].op=='Q') query(command[i].x,command[i].k);
            else if(command[i].op=='C') change_weight(command[i].x,command[i].k);
        }
        printf("Case %d: %.6lf\n",++cas,query_tot/(double)query_cnt);
    }
    return 0;
}