Time limit: 3.000 second
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:
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.
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.
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.
There are multiple test cases in the input file. Each case starts with two integers N and M (1
N2 * 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 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.
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
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
Case 1: 25.000000 Case 2: 16.666667
有关数据结构有Treap树，并查集，同时离线算法的设计。
/* * @author Panoss */ #include<iostream> #include<cstdio> #include<cstring> #include<string> #include<algorithm> #include<vector> #include<ctime> #include<stack> #include<queue> #include<list> using namespace std; #define DBG 0 #define fori(i,a,b) for(int i = (a); i < (b); i++) #define forie(i,a,b) for(int i = (a); i <= (b); i++) #define ford(i,a,b) for(int i = (a); i > (b); i--) #define forde(i,a,b) for(int i = (a); i >= (b); i--) #define forls(i,a,b,n) for(int i = (a); i != (b); i = n[i]) #define mset(a,v) memset(a, v, sizeof(a)) #define mcpy(a,b) memcpy(a, b, sizeof(a)) #define dout DBG && cerr << __LINE__ << " >>| " #define checkv(x) dout << #x"=" << (x) << " | "<<endl #define checka(array,a,b) if(DBG) { \ dout << #array"[] | " << endl; forie(i, a, b) cerr << "[" << i << "]=" << array[i] << " |" << ((i - (a)+1) % 5 ? " " : "\n"); if (((b)-(a)+1) % 5) cerr << endl; } #define redata(T, x) T x; cin >> x #define MIN_LD -2147483648 #define MAX_LD 2147483647 #define MIN_LLD -9223372036854775808 #define MAX_LLD 9223372036854775807 #define MAX_INF 18446744073709551615 inline int reint() { int d; scanf("%d", &d); return d; } inline long relong() { long l; scanf("%ld", &l); return l; } inline char rechar() { scanf(" "); return getchar(); } inline double redouble() { double d; scanf("%lf", &d); return d; } inline string restring() { string s; cin >> s; return s; } struct Treap_node { int value, priority, s; ///键值,优先级,总结点数(从根(包括根)开始往下算的结点个数) struct Treap_node * child[2]; Treap_node(int v): value(v) { child[0] = child[1] = NULL; priority = rand(); s = 1; } bool operator < (const Treap_node & X) const { return priority < X.priority; } int compare(int v) { if(v == value) return -1; return v < value ? 0 : 1; } void updata() { s = 1; if(child[0]) s += child[0]->s; if(child[1]) s += child[1]->s; } }; void Rotate(Treap_node* &root, int d) ///0左旋(逆时针),1右旋(顺时针) { Treap_node * k = root->child[d^1]; root->child[d^1] = k->child[d]; k->child[d] = root; root->updata(); k->updata(); root = k; } void Insert_node(Treap_node* &root, int v) { if(!root) { root = new Treap_node(v); } else { int d = (v < root->value ? 0 : 1); ///若允许有相同结点 ///int d = root->compare(v); ///若不允许有相同结点 Insert_node(root->child[d], v); if(root->child[d] > root) Rotate(root, d^1); } root->updata(); } void Remove_node(Treap_node * &root, int v) { int d = root->compare(v); if(d == -1) { Treap_node * temp = root; if(root->child[0] && root->child[1]) { int dir = root->child[0] > root->child[1] ? 1:0; Rotate(root, dir); Remove_node(root->child[dir], v); } else { if(!root->child[0]) root = root->child[1]; else root = root->child[0]; delete temp; } } else { Remove_node(root->child[d], v); } if(root) root->updata(); } bool Find_node(Treap_node * root, int v) { while(root) { int d = root->compare(v); if(d == -1) return true; else root = root->child[d]; } return false; } const int MAXN = 20000 + 10; const int MAXM = 60000 + 10; const int MAXC = 400000 + 10; struct Command { char op; int x, p; ///p = k or v }; Command cmd[MAXC]; int f[MAXN]; int Find(int x) {return f[x] == x? x : f[x] = Find(f[x]);} int from[MAXM], to[MAXM], weight[MAXN]; bool removed[MAXM]; Treap_node * Tree[MAXN]; int K_th(Treap_node * root, int k) { if(!root || k <= 0 || k > root->s) return 0; int s = (root->child[1]==NULL?0:root->child[1]->s); if(k == s+1) return root->value; else if(k <= s) return K_th(root->child[1],k); else return K_th(root->child[0],k-s-1); } void Merge_Tree(Treap_node* &from_root, Treap_node * &to_root) { if(from_root->child[0]) Merge_Tree(from_root->child[0],to_root); if(from_root->child[1]) Merge_Tree(from_root->child[1],to_root); Insert_node(to_root,from_root->value); delete from_root; from_root = NULL; } void Remove_Tree(Treap_node * &root) { if(root->child[0]) Remove_Tree(root->child[0]); if(root->child[1]) Remove_Tree(root->child[1]); delete root; root = NULL; } void Add_Edge(int e) { int u = Find(from[e]), v = Find(to[e]); if(u!=v) { if(Tree[u]->s < Tree[v]->s) { f[u] = v; Merge_Tree(Tree[u],Tree[v]); } else { f[v] = u; Merge_Tree(Tree[v],Tree[u]); } } } int query_cnt; long long query_tot; void Query(int x, int k) { query_cnt ++; query_tot += K_th(Tree[Find(x)], k); } void Change_weight(int x, int v) { int u = Find(x); Remove_node(Tree[u], weight[x]); Insert_node(Tree[u], v); weight[x] = v;/// } int main() { int Case = 0; int n,m; while(scanf("%d%d",&n,&m)==2&&(n+m)) { Case ++; forie(i,1,n) scanf("%d",&weight[i]); forie(i,1,m) scanf("%d%d",&from[i],&to[i]); ///Init mset(removed,false); forie(i,1,n) f[i] = i; query_cnt = query_tot = 0; int c = 0; for(;;) { char op; int x , p = 0, v = 0; scanf(" %c",&op); if(op == ‘E‘) break; scanf("%d",&x); if(op == ‘D‘) removed[x] = true; if(op == ‘Q‘) scanf("%d",&p); if(op == ‘C‘) { scanf("%d",&v); p = weight[x]; weight[x] = v; } cmd[++c] = (Command){op,x,p}; } forie(i,1,n) { if(Tree[i]) Remove_Tree(Tree[i]); Tree[i] = new Treap_node(weight[i]); } forie(i,1,m) if(!removed[i]) Add_Edge(i); forde(i,c,1) { if(cmd[i].op == ‘D‘) Add_Edge(cmd[i].x); if(cmd[i].op == ‘Q‘) Query(cmd[i].x,cmd[i].p); if(cmd[i].op == ‘C‘) Change_weight(cmd[i].x,cmd[i].p); } printf("Case %d: %.6lf\n",Case,query_tot/(double)query_cnt); } return 0; }
原文：http://www.cnblogs.com/Panoss/p/3810243.html