LA 5031 Graph and Queries

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题目链接：https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=3032

n(n<=2e4)个顶点m(m<=6e4)条边,每个顶点有个权值val_i, 然后有Q(Q<=5e5)次操作.

操作分为三类:

D x : 删除第x条边

Q x k : 查询与节点x关联的所有顶点中第k大

C x V : 将节点x的权值更改为V

输出查询的均值 /sum { Query_val } / Query_num

解题思

离线算法

对于删除，可以通过将所有操作读入后，从后往前处理。把删除边转换成插入边。

对于查询第k大顶点，我们可以使用 treap维护的名次树 kth来实现

对于修改操作，我们先将原来的值删除，然后再插入新值。因为我们使用离线逆向处理，则修改操作也会逆向。

关于名次树，只是对于 treap上增加了一个 size（其子节点的数量和+1）然后来实现求第K大 or 小.

关于Treap：Treap是一颗拥有键值v和优先级r两种权值的树。对于键值而言，这棵树是排序树；对于优先级而言，这棵树是大根堆。

不难证明，如果每个节点的优先级事先给定且不互相等，这棵树的形态也唯一确定了。

关于Treap的模板：

struct Node{    Node * ch[2];    int r;//优先级    int v;//值    int s;//节点个数    Node(int _v)    {        v = _v;        ch[0] = ch[1] = NULL;        r = rand();        s = 1;    }    //根据优先级比较节点    bool operator < (const Node & rhs) const    {        return r < rhs.r;    }    //比较值确定插入的方向    int cmp(int x) const    {        if(v == x) 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;    }};//d=0向左转，d=1向右转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] > 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 o = o->ch[0];            delete u;        }    }    else    {        remove(o->ch[d],x);    }    if(o!=NULL) o->maintain();}

Treap可以用来实现名次树：

在每次树中，每一个节点有一个附加size,表示以它为根的子树的总节点数。

Kth（k）:找出第K小元素：

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);}

本题代码：

#include <iostream>#include <stdio.h>#include <stdlib.h>#include <string.h>#include <math.h>#include <map>#include <queue>#include <algorithm>using namespace std;struct Node{    Node * ch[2];    int r;//优先级    int v;//值    int s;//节点个数    Node(int _v)    {        v = _v;        ch[0] = ch[1] = NULL;        r = rand();        s = 1;    }    //根据优先级比较节点    bool operator < (const Node & rhs) const    {        return r < rhs.r;    }    //比较值确定插入的方向    int cmp(int x) const    {        if(v == x) 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;    }};//d=0向左转，d=1向右转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] > 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 o = o->ch[0];            delete u;        }    }    else    {        remove(o->ch[d],x);    }    if(o!=NULL) o->maintain();}#define Maxn 20005#define Maxm 60005#define Maxc 500005struct Command{    char type;    int x,p;    Command(){}    Command(char _type,int _x,int _p)    {        type = _type;        x = _x;        p = _p;    }};Command commands[Maxc];//边的起始、终止节点编号int from[Maxm],to[Maxm];//已经删除的边号int removed[Maxm];//每个节点的值int weight[Maxn];//父亲节点int pa[Maxn];//名次树的根节点Node * root[Maxn];//查询的次数int query_cnt;//查询得到的值的和long long query_tot;int findset(int x){    if(x == pa[x]) return x;    pa[x] = findset(pa[x]);    return pa[x];}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;}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;}void add_edge(int x){    int u = findset(from[x]);    int v = findset(to[x]);    if(u!=v)    {        if(root[u]->s > root[v]->s) {pa[v] = u;mergeto(root[v],root[u]);}        else{pa[u] = v;mergeto(root[u],root[v]);}    }}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 query(int x,int k){    query_cnt++;    query_tot += kth(root[findset(x)],k);    //printf()}void change_weight(int x,int v){    int u = findset(x);    remove(root[u],weight[x]);    insert(root[u],v);    weight[x] = v;}int main(){#ifndef ONLINE_JUDGE    freopen("in.txt","r",stdin);#endif    int n,m;    int x,p,v;    char type;    int cas = 0;    while(scanf(" %d %d",&n,&m)!=EOF && n!=0)    {        cas++;        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));        int c = 0;        while(scanf(" %c",&type)!=EOF && type != 'E')        {            scanf(" %d",&x);            if(type == 'C')            {                scanf(" %d",&v);                p = weight[x];                weight[x] = v;            }            if(type == 'D') removed[x] = 1;            if(type == 'Q') scanf(" %d",&p);            Command a(type,x,p);            commands[c++] = a;        }        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);        //反向操作        query_cnt = query_tot = 0;        for(int i=c-1;i>=0;i--)        {            if(commands[i].type == 'D') add_edge(commands[i].x);            if(commands[i].type == 'Q') query(commands[i].x,commands[i].p);            if(commands[i].type == 'C') change_weight(commands[i].x,commands[i].p);        }        printf("Case %d: %.6lf\n",cas,query_tot/(double)query_cnt);    }    return 0;}