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 [-10 6, 10 6].
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.
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 [-10 6, 10 6].
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 * 10
4, 0 <= M <= 6 * 10
4), the number of vertexes in the graph. The next N lines describes the initial weight of each vertex (-106 <= weight[i] <= 10
6). 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 * 10
5], and there will be no more than 2 * 10
5 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.
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: 16.666667HintFor 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;
}