uvalive 5031

L - Graph and Queries

Time Limit:3000MS    Memory Limit:0KB    64bit IO Format:%lld & %llu

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Description

You are given an undirected graph with N vertexes andM 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 vertexX (including X itself).

   The format is [Q X K], where X is an integer from 1 toN, indicating the id of the vertex, and you may assume thatK will always fit into a 32-bit signed integer. In caseK 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 toN, 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 integersN and M (1N2* 10⁴, 0M6* 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 toN. 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 [QX 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 [CX 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 50Q 1 1E3 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

一个图，一堆操作，计算多次查询的第k大平均值。

离线读入命令，逆序操作，平衡树启发式合并，

代码：

/* ***********************************************Author :xianxingwuguanCreated Time :2014/3/4 12:48:23File Name :G.cpp************************************************ */#pragma comment(linker, "/STACK:102400000,102400000")#include <stdio.h>#include <iostream>#include <algorithm>#include <sstream>#include <stdlib.h>#include <string.h>#include <limits.h>#include <string>#include <time.h>#include <math.h>#include <queue>#include <stack>#include <set>#include <map>using namespace std;#define INF 0x3f3f3f3f#define eps 1e-8#define pi acos(-1.0)typedef long long ll;struct Node{Node *ch[2];int r;int v;int s;Node(int v):v(v){ch[0]=ch[1]=NULL;r=rand();s=1;}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;}void maintain(){s=1;if(ch[0]!=NULL)s+=ch[0]->s;if(ch[1]!=NULL)s+=ch[1]->s;}};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();}const int maxc=500040;struct Command{char type;int x,p;}command[maxc];const int maxn=200010;const int maxm=100100;int n,m,weight[maxn],from[maxm],to[maxm],removed[maxn];int pa[maxn];int find(int x){if(pa[x]!=x)pa[x]=find(pa[x]);return pa[x];}Node *root[maxn];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);return kth(o->ch[0],k-s-1);}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=find(from[x]);int v=find(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;ll query_tot;void query(int x,int k){query_cnt++;query_tot+=kth(root[find(x)],k);}void change_weight(int x,int v){int u=find(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));    // 读命令        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;            }            command[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(command[i].type == 'D')                add_edge(command[i].x);            if(command[i].type == 'Q')                query(command[i].x, command[i].p);            if(command[i].type == 'C')                change_weight(command[i].x, command[i].p);        }        printf("Case %d: %.6lf\n", ++kase, query_tot / (double)query_cnt);    }    return 0;}