LA 5031 Graph and Queries （【名次树(treap)】+【并查集】+【离线算法】）

题目链接：https://cn.vjudge.net/problem/UVALive-5031

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.

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

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

【中文题意】

有一张n个节点m条边的无向图，每个结点都有一个整数权值。你的任务是执行一系列操作。操作有如下三种：

1.D X(1<=X <=m) 删除ID为X的边，输入保证每条边至多被删除1次。

2.Q X k(1<=X <=n)，计算与节点X连通的结点中（包括X本身），第k大的权值，如果不存在，返回0 。

3.C X V (1<=x <=n)把结点X的权值改为V

操作序列结束的标志位单个字母E，节点编号为1~n,边编号为1~m。

输入格式

输入第一行为两个整数n和m（1<=n<=20000,0<=m<=60000），以下n行每行有一个绝对值不超过10^6 的整数，即各结点的初始权值，以下m行每行有两个整数，即一条边的两个端点。接下来是每条指令，以单个字母E结尾。保证Q和C的指令均不超过200000条，输入结束标志位n=m=0。

输出格式

对于每组数据，输出所有Q指令的计算结果的平均值，精确到小数点后6位。

【分析】

本题只需要设计离线算法，因此可以把操作的顺序反过来处理，先读入所有操作，执行所有的D操作得到最终的图，然后按照逆序把这些边逐步插入，并在恰当的时机执行Q和C操作。用一棵名次树维护一个联通分量中的点权，则C操作对应名次树中一次修改操作（可以用一次删除加一次插入来实现），Q操作对应kth操作，而执行D操作时，如果两个端点已经是一个同一连通分量则无影响，否则将两个端点对应的名次树合并。

【AC代码】

#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
using namespace std;

struct Node
{
Node *ch[2];//左右子树
int r;//随机优先级
int v;//该点的值
int s;//该点的节点总数
Node(int v):v(v)
{
ch[0]=ch[1]=NULL;//左右子树初始化为空
r=rand();//优先级随机
s=1;//只有本身一个结点
}
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;//0代表左边，1代表右边
}
void maintain()//更新结点域
{
s=1;
if(ch[0]!=NULL)s+=ch[0]->s;
if(ch[1]!=NULL)s+=ch[1]->s;
}
};
void rotate(Node* &o,int d)//对o结点进行旋转，0代表左旋，1代表右旋，旋转后更新结点域
{
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)//插入一个值为x的结点
{
if(o==NULL)o=new Node(x);
else
{
int d=(x<o->v?0:1);//x< o->v 时d=0该节点被插入左边，否则d=1插入右边
insert(o->ch[d],x);
if(o->ch[d]>o)rotate(o,d^1);//如果被插入后优先级比父节点大，要发生旋转
}
o->maintain();
}
void remove(Node* &o,int x)//删除一个值为x的节点
{
int d=o->cmp(x);
if(d==-1)//d==1时代表代表需删除的节点为当前节点
{
Node* u=o;
if(o->ch[0]!=NULL&&o->ch[1]!=NULL)//左右孩子非空
{
int d2=(o->ch[0] > o->ch[1] ?1:0);//左孩子与右孩子谁的优先级比较高
rotate(o,d2);//谁的高向另外的方向旋转
remove(o->ch[d2],x);//旋转后删除x节点
}
else
{
if(o->ch[0]==NULL)o=o->ch[1];//左孩子为空，o指向右孩子
else o=o->ch[0];//否则指向左孩子
delete u;
}
}
else
remove(o->ch[d],x);
if(o!=NULL)o->maintain();//删除后o有孩子的话更新结点域
}

const int maxc = 500000 + 10;
struct Command
{
char type;
int x,p;
}commands[maxc];//记录要执行的命令

const int maxn = 20000 + 10;//点数最大
const int maxm = 60000 + 10;//边数最大
int n,m,weight[maxn],from[maxm],to[maxm],removed[maxm];

//利用并查集维护联通关系
int pa[maxn];
int findset(int x)
{
return pa[x]!=x ? pa[x]=findset(pa[x]) : x;
}

Node *root[maxn];//Treap

int kth(Node* o,int k)//查找以o为根节点的子树中的第k大
{
if(o==NULL || k <= 0 ||k > o->s)return 0;//不合法情况
int s = (o->ch[1] == NULL ? 0 : o->ch[1]->s);
if(k==s+1)return o->v;//正好第k大为当前节点
else if(k<=s)
{
return kth(o->ch[1],k);//第k大在右子树上
}
else return kth(o->ch[0],k-s-1);//第k大在左子树上
}

void mergeto(Node* &src,Node* &dest)//合并两棵子树
{
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)//删除根节点为x的子树
{
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)//增加第x条边
{
int u=findset(from[x]),v=findset(to[x]);
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]);
}
}
}

int query_cnt;
long long query_tot;
void query(int x,int k)//查询与x联通的所有结点中的第k大
{
query_cnt++;
query_tot+=kth(root[findset(x)],k);
}

void change_weight(int x,int v)//把结点x的值改为v 操作为：先删除，后添加
{
int u=findset(x);
remove(root[u],weight[x]);
insert(root[u],v);
weight[x]=v;
}

int main()
{
int kase=0;
while(scanf("%d%d",&n,&m)==2 &&n)
{
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;
for(;;)
{
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};
}

//最终的图
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_tot = query_cnt=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",++kase,query_tot/(double)query_cnt);
}
return 0;
}