HDU 3726 Graph and Queries （离线处理+splay tree）

Graph and Queries

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1467    Accepted Submission(s): 301

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

 

题目首先给出了N个点，M条边的图。每个点有一个对应的权值。

然后下面由三种操作：

1： D X :删掉第X条边。

2： Q  X  K ：查询和X相连的点中 第K大的点的值（K=1表示最大，K=2表示第二大，。。。。）

3： C X V ：将点X的权值修改为V

离线处理。

点的权值的变化使用邻接表存下来。

离线后删边当成加边，就是合并。

合并的时候只能把点的个数少的一个个插入

/* ***********************************************
Author        :kuangbin
Created Time  :2013/8/28 23:05:49
File Name     :F:2013ACM练习专题学习splay_tree_2HDU3726.cpp
************************************************ */

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;

#define Key_value ch[ch[root][1]][0]
const int MAXN = 20010;
int pre[MAXN],ch[MAXN][2],key[MAXN],size[MAXN];
int root;

void NewNode(int &r,int father,int loc,int k)
{
    r = loc;
    pre[r] = father;
    ch[r][0] = ch[r][1] = 0;
    key[r] = k;
    size[r] = 1;
}
void push_up(int r)
{
    size[r] = size[ch[r][0]] + size[ch[r][1]] + 1;
}

void Init()
{
    root = 0;
    ch[root][0] = ch[root][1] = key[root] = size[root] = 0;
    pre[root] = 0;
}

void Rotate(int x,int kind)
{
    int y = pre[x];
    ch[y][!kind] = ch[x][kind];
    pre[ch[x][kind]] = y;
    if(pre[y])
        ch[pre[y]][ch[pre[y]][1]==y] = x;
    pre[x] = pre[y];
    ch[x][kind] = y;
    pre[y] = x;
    push_up(y);
}
void Splay(int r,int goal)
{
    while(pre[r] != goal)
    {
        if(pre[pre[r]] == goal)
            Rotate(r,ch[pre[r]][0] == r);
        else
        {
            int y = pre[r];
            int kind = ch[pre[y]][0]==y;
            if(ch[y][kind] == r)
            {
                Rotate(r,!kind);
                Rotate(r,kind);
            }
            else
            {
                Rotate(y,kind);
                Rotate(r,kind);
            }
        }
    }
    push_up(r);
    if(goal == 0) root = r;
}
int Get_kth(int r,int k)
{
    int t = size[ch[r][0]] + 1;
    if(t == k)return r;
    if(t > k)return Get_kth(ch[r][0],k);
    else return Get_kth(ch[r][1],k-t);
}

void Insert(int loc,int k)
{
    int r = root;
    if(r == 0)
    {
        NewNode(root,0,loc,k);
        return;
    }
    while(ch[r][key[r]<k])
        r = ch[r][key[r]<k];
    NewNode(ch[r][key[r]<k],r,loc,k);
    Splay(ch[r][key[r]<k],0);
}


struct Edge
{
    int u,v;
}edge[60010];
bool used[60010];

//把修改的值建成邻接表
int to[400010];
int next[400010];
int head[20010];
int tot;

void add_value(int x,int v)
{
    to[tot] = v;
    next[tot] = head[x];
    head[x] = tot++;
}

struct Query
{
    char op[5];
    int x,y;
}query[400010];
int q_num;

int F[20010];

int find(int x)
{
    if(F[x] == -1)return x;
    else return F[x] = find(F[x]);
}

void erase(int r)
{
    if(!r)return;
    erase(ch[r][0]);
    erase(ch[r][1]);
    Insert(r,to[head[r]]);
}
int Get_Min(int r)
{
    while(ch[r][0])
    {
        r = ch[r][0];
    }
    return r;
}

//删除根结点
void Delete()
{
    if(ch[root][0] == 0 || ch[root][1] == 0)
    {
        root = ch[root][0] + ch[root][1];
        pre[root] = 0;
        return;
    }
    int k = Get_Min(ch[root][1]);
    Splay(k,root);
    Key_value = ch[root][0];
    root = ch[root][1];
    pre[ch[root][0]] = root;
    pre[root] = 0;
    push_up(root);
}
void bing(int x,int y)
{
    int t1 = find(x),t2 = find(y);
    if(t1 == t2)return;
    F[t1] = t2;
    Splay(t1,0);
    Splay(t2,0);
    if(size[t1] > size[t2])
        swap(t1,t2);
    root = t2;
    erase(t1);
}
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int N,M;
    int iCase = 0;
    while(scanf("%d%d",&N,&M) == 2 )
    {
        if(N == 0 && M == 0) break;
        iCase++;
        memset(head,-1,sizeof(head));
        tot = 0;
        int v;
        for(int i = 1;i <= N;i++)
        {
            scanf("%d",&v);
            add_value(i,v);
        }
        for(int i = 1;i <= M;i++)
        {
            scanf("%d%d",&edge[i].u,&edge[i].v);
        }
        q_num = 0;
        memset(used,false,sizeof(used));
        while(scanf("%s",&query[q_num].op) == 1)
        {
            if(query[q_num].op[0] == 'E')break;
            if(query[q_num].op[0] == 'D')
            {
                scanf("%d",&query[q_num].x);
                used[query[q_num].x] = true;
            }
            else if(query[q_num].op[0] == 'Q')
                scanf("%d%d",&query[q_num].x,&query[q_num].y);
            else if(query[q_num].op[0] == 'C')
            {
                scanf("%d%d",&query[q_num].x,&query[q_num].y);
                add_value(query[q_num].x,query[q_num].y);
            }
            q_num++;
        }
        memset(F,-1,sizeof(F));
        for(int i = 1;i <= N;i++)
            NewNode(root,0,i,to[head[i]]);

        for(int i = 1;i <= M;i++)
            if(!used[i])
                bing(edge[i].u,edge[i].v);

        double ans = 0;
        int cnt = 0;
        
        for(int i = q_num-1; i>= 0 ;i--)
        {
            if(query[i].op[0] == 'Q')
            {
                Splay(query[i].x,0);
                if(size[root] < query[i].y || query[i].y <= 0)
                {
                    cnt++;
                    continue;
                }
                ans += key[Get_kth(root,size[root] - query[i].y + 1)];
                cnt++;
            }
            else if(query[i].op[0] == 'D')
            {
                int tmp = query[i].x;
                bing(edge[tmp].u,edge[tmp].v);
            }
            else if(query[i].op[0] == 'C')
            {
                Splay(query[i].x,0);
                Delete();
                head[query[i].x] = next[head[query[i].x]];
                Insert(query[i].x,to[head[query[i].x]]);
            }
        }    
        if(cnt == 0)cnt = 1;
        printf("Case %d: %.6lf
",iCase,ans/cnt);
    }
    return 0;
}