hdu 3726 Graph and Queries 名次树

Graph and Queries

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1333    Accepted Submission(s): 267

Problem Description

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:
1)  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.
2)  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.
3)  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 <= N <= 2 * 10⁴, 0 <= M <= 6 * 10⁴), the number of vertexes in the graph. The next N lines describes the initial weight of each vertex (-106 <= weight[i] <= 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.

 

Sample Input

3 3
10
20
30
1 2
2 3
1 3
D 3
Q 1 2
Q 2 1
D 2
Q 3 2
C 1 50
Q 1 1
E

3 3
10
20
20
1 2
2 3
1 3
Q 1 1
Q 1 2
Q 1 3
E

0 0

 

Sample Output

Case 1: 25.000000
Case 2: 16.666667

Hint
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

 

 

Source

2010 Asia Tianjin Regional Contest

 

Recommend

zhouzeyong

 

----------

将操作反序处理，由删除边变为添加边。

D 合并边X顶点的集合

C 一次删除一次插入

Q Treap询问Kth

----------

#include <iostream>
#include <ctime>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;

struct Node{
    Node* ch[2];//左右子树
    int fix;//优先级。数值越大，优先级越高
    int key;
    int size;//以它为根的子树的总结点数
    bool operator<(const Node& rhs) const {
        return fix<rhs.fix;
    }
    int cmp(int x) const{
        if (x==key) return -1;
        return x<key?0:1;
    }
    //名次树
    void maintain(){
        size=1;
        if (ch[0]!=NULL) size+=ch[0]->size;
        if (ch[1]!=NULL) size+=ch[1]->size;
    }
};

struct Treap{
    Node* root;
    Treap(){
        srand(time(0));
        root=NULL;
    }
    void removetree(Node* &t){
        if (t->ch[0]!=NULL) removetree(t->ch[0]);
        if (t->ch[1]!=NULL) removetree(t->ch[1]);
        delete t;
        t=NULL;
    }
    void clear(){
        srand(time(0));
        removetree(root);
    }
    Node* newNode(int v){
        Node* t=new Node;
        t->key=v;
        t->ch[0]=t->ch[1]=NULL;
        t->fix=rand();
        t->size=1;
        return t;
    }
    //d=0代表左旋,d=1代表右旋
    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;
    }
    //在以o为根的子树中插入键值x,修改o
    void insert(Node* &o,int x){
        if (o==NULL) o=newNode(x);
        else{
            int d=o->cmp(x);
            if (d==-1) d=1;
            insert(o->ch[d],x);
            if (o->ch[d]>o) 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]>o->ch[1]?1:0);
                rotate(o,d2);
                remove(o->ch[d2],x);
            }else{
                if (o->ch[0]==NULL) o=o->ch[1];
                else if (o->ch[1]==NULL) o=o->ch[0];
                delete u;
            }
        }
        else remove(o->ch[d],x);
        if (o!=NULL) o->maintain();
    }
    bool find(Node* o,int x){
        while (o!=NULL){
            int d=o->cmp(x);
            if (d==-1) return 1;
            else o=o->ch[d];
        }
        return 0;
    }
    //第k大的值
    int kth(Node* o,int k){
        if (o==NULL||k<=0||k>o->size) return 0;
        int s=(o->ch[1]==NULL?0:o->ch[1]->size);
        if (k==s+1) return o->key;
        else if (k<=s) return kth(o->ch[1],k);
        else return kth(o->ch[0],k-s-1);
    }
    void merge(Node* &src){
        if (src->ch[0]!=NULL) merge(src->ch[0]);
        if (src->ch[1]!=NULL) merge(src->ch[1]);
        insert(root,src->key);
        delete src;
        src=NULL;
    }
};

const int maxc=500000+10;
const int maxn=20000+10;
const int maxm=60000+10;
struct Command{
    char type;
    int x,p;
    Command(char _type=0,int _x=0,int _p=0){
        type=_type;
        x=_x;
        p=_p;
    }
}commands[maxc];
int n,m,weight[maxn],from[maxm],to[maxm],removed[maxm];

//并查集相关
int pa[maxn];
void makeset(int n){
    for (int i=0;i<=n;i++) pa[i]=i;
}
int findset(int x){
    if (x!=pa[x]) pa[x]=findset(pa[x]);
    return pa[x];
}
void unionset(int x,int y){
    x=findset(x);
    y=findset(y);
    if (x!=y) pa[x]=y;
}
//名次树相关
Treap T[maxn];
//D操作
void D_addedge(int x)
{
    int u=findset(from[x]);
    int v=findset(to[x]);
    if (u!=v)
    {
        if (T[u].root->size<T[v].root->size)
        {
            T[v].merge(T[u].root);
            pa[u]=v;
        }
        else
        {
            T[u].merge(T[v].root);
            pa[v]=u;
        }
    }
}
//Q操作
int query_cnt;
long long query_tot;
void Q_query(int x,int k)
{
    query_cnt++;
    query_tot+=T[findset(x)].kth(T[findset(x)].root,k);
}
//C操作
void C_change(int x,int v)
{
    int u=findset(x);
    T[u].remove(T[u].root,weight[x]);
    T[u].insert(T[u].root,v);
    weight[x]=v;
}

int main()
{
    int cas=0;
    while (~scanf("%d%d",&n,&m))
    {
        if (n==0&&m==0) break;
        for (int i=1;i<=n;i++) scanf("%d",&weight[i]);
        for (int i=1;i<=m;i++) scanf("%d%d",&from[i],&to[i]);
        memset(removed,0,sizeof(removed));
        //读命令
        int c=0;
        while (1){
            char type;
            int x,p=0,v=0;
            scanf(" %c",&type);
            if (type=='E') break;
            scanf("%d",&x);
            if (type=='D') removed[x]=1;
            if (type=='Q') scanf("%d",&p);
            if (type=='C'){
                scanf("%d",&v);
                p=weight[x];
                weight[x]=v;
            }
            commands[c++]=Command(type,x,p);
        }
        makeset(n);
        for (int i=1;i<=n;i++)
        {
            if (T[i].root!=NULL) T[i].clear();
            T[i].insert(T[i].root,weight[i]);
        }
        for (int i=1;i<=m;i++)
        {
            if (!removed[i]) D_addedge(i);
        }
        //反向操作
        query_cnt=query_tot=0;
        for (int i=c-1;i>=0;i--)
        {
            if (commands[i].type=='D') D_addedge(commands[i].x);
            if (commands[i].type=='Q') Q_query(commands[i].x,commands[i].p);
            if (commands[i].type=='C') C_change(commands[i].x,commands[i].p);
        }
        printf("Case %d: %0.6f\n",++cas,query_tot/(double)query_cnt);
    }
    return 0;
}