HDU 3726 Graph and Queries（平衡二叉树）（2010 Asia Tianjin Regional Contest）

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.
 
题目大意：给一个n个点m条边的无向图，有三种询问，分别为删边、询问某子集内第k大权、修改某点权值，问数次询问后，第二种询问的值的平均数
思路：用treap树维护一个强联通分量。离线处理所有询问，先删掉图中将会被删掉的边，从后往前询问，修改权值的询问也要稍作处理。
PS：因为打错一个字母RE了半天……
 
  1 #include <cstdlib>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <algorithm>
  5 using namespace std;
  6 
  7 const int MAXC = 500010;
  8 const int MAXN = 20010;
  9 const int MAXM = 60010;
 10 
 11 int key[MAXN], weight[MAXN], child[MAXN][2], size[MAXN];
 12 int stk[MAXN], top, poi_cnt;//not use point
 13 
 14 inline int newNode(int k) {
 15     int x = (top ? stk[top--] : ++poi_cnt);
 16     key[x] = k;
 17     size[x] = 1;
 18     weight[x] = rand();
 19     child[x][0] = child[x][1] = 0;
 20     return x;
 21 }
 22 
 23 inline void update(int &x) {
 24     size[x] = size[child[x][0]] + size[child[x][1]] + 1;//size[0]=0
 25 }
 26 
 27 inline void rotate(int &x, int t) {
 28     int y = child[x][t];
 29     child[x][t] = child[y][t ^ 1];
 30     child[y][t ^ 1] = x;
 31     update(x); update(y);
 32     x = y;
 33 }
 34 
 35 void insert(int &x, int k) {
 36     if (x == 0) x = newNode(k);
 37     else {
 38         int t = (key[x] < k);
 39         insert(child[x][t], k);
 40         if (weight[child[x][t]] < weight[x]) rotate(x, t);
 41     }
 42     update(x);
 43 }
 44 
 45 void remove(int &x, int k) {
 46     if(key[x] == k) {
 47         if(child[x][0] && child[x][1]) {
 48             int t = weight[child[x][0]] < weight[child[x][1]];
 49             rotate(x, t); remove(child[x][t ^ 1], k);
 50         }
 51         else {
 52             stk[++top] = x;
 53             x = child[x][0] + child[x][1];
 54         }
 55     }
 56     else remove(child[x][key[x] < k], k);
 57     if(x > 0) update(x);
 58 }
 59 
 60 struct Command {
 61     char type;
 62     int x, p;
 63 } commands[MAXC];
 64 
 65 int n, m, value[MAXN], from[MAXM], to[MAXM], removed[MAXM];
 66 
 67 int fa[MAXN];
 68 int getfather(int x) {
 69     if(fa[x] == x) return x;
 70     else return fa[x] = getfather(fa[x]);
 71 }
 72 
 73 int root[MAXN];
 74 
 75 void mergeto(int &x, int &y) {
 76     if(child[x][0]) mergeto(child[x][0], y);
 77     if(child[x][1]) mergeto(child[x][1], y);
 78     insert(y, key[x]);
 79     stk[++top] = x;
 80 }
 81 
 82 inline void addEdge(int x) {
 83     int u = getfather(from[x]), v = getfather(to[x]);
 84     if(u != v) {
 85         if(size[root[u]] > size[root[v]]) swap(u, v);
 86         fa[u] = v; mergeto(root[u], root[v]);
 87     }
 88 }
 89 
 90 int kth(int &x, int k) {
 91     if(x == 0 || k <= 0 || k > size[x]) return 0;
 92     int s = 0;
 93     if(child[x][1]) s = size[child[x][1]];
 94     if(k == s + 1) return key[x];
 95     if(k <= s) return kth(child[x][1], k);
 96     return kth(child[x][0], k - s - 1);
 97 }
 98 
 99 int query_cnt;
100 long long query_tot;
101 
102 void query(int x, int k) {
103     ++query_cnt;
104     query_tot += kth(root[getfather(x)], k);
105 }
106 
107 inline void change_value(int x, int v) {
108     int u = getfather(x);
109     remove(root[u], value[x]);
110     insert(root[u], value[x] = v);
111 }
112 
113 int main() {
114     int kase = 0;
115     size[0] = 0;
116     while(scanf("%d%d", &n, &m) != EOF && n) {
117         for(int i = 1; i <= n; ++i) scanf("%d", &value[i]);
118         for(int i = 1; i <= m; ++i) scanf("%d%d", &from[i], &to[i]);
119         memset(removed, 0, sizeof(removed));
120 
121         int c = 0;
122         while(true) {
123             char type;
124             int x, p = 0, v = 0;
125             scanf(" %c", &type);
126             if(type == 'E') break;
127             scanf("%d", &x);
128             if(type == 'D') removed[x] = 1;
129             if(type == 'Q') scanf("%d", &p);
130             if(type == 'C') {
131                 scanf("%d", &v);
132                 p = value[x];
133                 value[x] = v;
134             }
135             commands[c++] = (Command) {type, x, p};
136         }
137 
138         top = poi_cnt = 0;
139         for(int i = 1; i <= n; ++i) {
140             fa[i] = i;
141             root[i] = newNode(value[i]);
142         }
143         for(int i = 1; i <= m; ++i) if(!removed[i]) addEdge(i);
144 
145         query_tot = query_cnt = 0;
146         for(int i = c - 1; i >= 0; --i) {
147             if(commands[i].type == 'D') addEdge(commands[i].x);
148             if(commands[i].type == 'Q') query(commands[i].x, commands[i].p);
149             if(commands[i].type == 'C') change_value(commands[i].x, commands[i].p);
150         }
151         printf("Case %d: %.6f\n", ++kase, query_tot/(double)query_cnt);
152     }
153     return 0;
154 }
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posted @ 2013-07-26 23:14  Oyking  阅读(391)  评论(0)  编辑  收藏  举报