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_2\HDU3726.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 = ;
 int pre[MAXN],ch[MAXN][],key[MAXN],size[MAXN];
 int root;

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

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

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

 void Insert(int loc,int k)
 {
     int r = root;
     if(r == )
     {
         NewNode(root,,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],);
 }

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

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

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

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

 int F[];

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

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

 //删除根结点
 void Delete()
 {
     if(ch[root][] ==  || ch[root][] == )
     {
         root = ch[root][] + ch[root][];
         pre[root] = ;
         return;
     }
     int k = Get_Min(ch[root][]);
     Splay(k,root);
     Key_value = ch[root][];
     root = ch[root][];
     pre[ch[root][]] = root;
     pre[root] = ;
     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,);
     Splay(t2,);
     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 = ;
     while(scanf("%d%d",&N,&M) ==  )
     {
         if(N ==  && M == ) break;
         iCase++;
         memset(head,-,sizeof(head));
         tot = ;
         int v;
         for(int i = ;i <= N;i++)
         {
             scanf("%d",&v);
             add_value(i,v);
         }
         for(int i = ;i <= M;i++)
         {
             scanf("%d%d",&edge[i].u,&edge[i].v);
         }
         q_num = ;
         memset(used,false,sizeof(used));
         while(scanf("%s",&query[q_num].op) == )
         {
             if(query[q_num].op[] == 'E')break;
             if(query[q_num].op[] == 'D')
             {
                 scanf("%d",&query[q_num].x);
                 used[query[q_num].x] = true;
             }
             else if(query[q_num].op[] == 'Q')
                 scanf("%d%d",&query[q_num].x,&query[q_num].y);
             else if(query[q_num].op[] == '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,-,sizeof(F));
         for(int i = ;i <= N;i++)
             NewNode(root,,i,to[head[i]]);

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

         double ans = ;
         int cnt = ;

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