题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3726
题意：n个点m条边的一张无向图，每个点有一个权值， 有3中操作。
D X 删除第X条边
Q X K 计算与X点相连所有点中第k大的权值
C X V把X的权值改为 V
输出 Q次询问的平均值
大白上的例题， 离线处理，把所有操作 反过来处理，，这样删边变成了 加边，，瞬间好处理多了。。细节也有很多。
1 #include <set>
2 #include <map>
3 #include <cmath>
4 #include <ctime>
5 #include <queue>
6 #include <stack>
7 #include <cstdio>
8 #include <string>
9 #include <vector>
10 #include <cstdlib>
11 #include <cstring>
12 #include <iostream>
13 #include <algorithm>
14 using namespace std;
15 typedef unsigned long long ull;
16 typedef long long ll;
17 const int inf = 0x3f3f3f3f;
18 const double eps = 1e-8;
19 const int maxn = 2e4+10;
20 int n, m, val[maxn];
21 int pa[maxn];
22 void init()
23 {
24 for (int i = 0; i <= n; i++)
25 {
26 pa[i] = i;
27 }
28 }
29 int find(int x)
30 {
31 return pa[x] = (pa[x] == x ? x : find(pa[x]));
32 }
33 struct Node
34 {
35 Node* ch[2];
36 int key, r, siz;
37 Node (int x)
38 {
39 key = x;
40 siz = 1;
41 ch[0] = ch[1] = NULL;
42 r = rand();
43 }
44 bool operator < (const Node &rhs)const
45 {
46 return r < rhs.r;
47 }
48 int cmp(int x)
49 {
50 if (x == key)
51 return -1;
52 return x < key ? 0 : 1;
53 }
54 void maintain()
55 {
56 siz = 1;
57 if (ch[0] != NULL)
58 siz += ch[0] -> siz;
59 if (ch[1] != NULL)
60 siz += ch[1] -> siz;
61 }
62 };
63 void rotate(Node* &root, int d)
64 {
65 Node* tmp = root -> ch[d^1];
66 root -> ch[d^1] = tmp -> ch[d];
67 tmp -> ch[d] = root;
68 root -> maintain();
69 tmp -> maintain();
70 root = tmp;
71 }
72 void insert(Node* &root, int x)
73 {
74 if (root == NULL)
75 root = new Node (x);
76 else
77 {
78 int d = x < root -> key ? 0 : 1;
79 insert (root ->ch[d], x);
80 if (root -> ch[d] -> r > root -> r)
81 rotate(root, d^1);
82
83 }
84 root -> maintain();
85 }
86 void dele(Node* &root, int x)
87 {
88 int d = root -> cmp(x);
89 if (d == -1)
90 {
91 Node* tmp = root;
92 if (root -> ch[0] != NULL && root -> ch[1] != NULL)
93 {
94 int d1 = (root -> ch[0] -> r > root -> ch[1] -> r ? 0 : 1);
95 rotate(root, d1^1);
96 dele(root -> ch[d1^1], x);
97 }
98 else
99 {
100 if (root -> ch[0] == NULL)
101 root = root -> ch[1];
102 else
103 root = root -> ch[0];
104 delete tmp;
105 }
106 }
107 else
108 dele(root->ch[d], x);
109 if (root != NULL)
110 root -> maintain();
111 }
112 int Get_kth(Node* root, int k)
113 {
114 if (root == NULL || k <= 0 || root -> siz < k)
115 return 0;
116 int s = root -> ch[1] == NULL ? 0 : root -> ch[1] -> siz;
117 if (s + 1 == k)
118 return root -> key;
119 if (s + 1 > k)
120 return Get_kth(root -> ch[1], k);
121 else
122 return Get_kth(root -> ch[0], k-s-1);
123
124 }
125 Node* root[maxn];
126 void merge_tree(Node* &src, Node* &fa)
127 {
128 if (src -> ch[0] != NULL)
129 merge_tree(src -> ch[0], fa);
130 if (src -> ch[1] != NULL)
131 merge_tree(src -> ch[1], fa);
132 insert(fa, src -> key);
133 delete src;
134 src = NULL;
135 }
136 void remove_tree(Node* &root)
137 {
138 if (root -> ch[0] != NULL)
139 remove_tree(root -> ch[0]);
140 if (root -> ch[1] != NULL)
141 remove_tree(root -> ch[1]);
142 delete root;
143 root = NULL;
144 }
145 struct
146 {
147 int x, y;
148 bool is_del;
149 } e[3 * maxn];
150 struct
151 {
152 int type, x, p;
153 } Q[maxn*25];
154 void add_edge(int idx)
155 {
156 int fx = find(e[idx].x);
157 int fy = find(e[idx].y);
158 if (fx != fy)
159 {
160 if (root[fx] -> siz > root[fy] -> siz)
161 {
162 pa[fy] = fx;
163 merge_tree(root[fy], root[fx]);
164 }
165 else
166 {
167 pa[fx] = fy;
168 merge_tree(root[fx], root[fy]);
169
170 }
171 }
172
173 }
174 int main(void)
175 {
176 #ifndef ONLINE_JUDGE
177 freopen("in.txt","r",stdin);
178 #endif
179 int cas = 1;
180 while (~ scanf ("%d%d",&n, &m))
181 {
182 if (n == 0 && m == 0)
183 break;
184 for (int i = 0; i < n; i++)
185 {
186 scanf ("%d", val + 1 + i);
187 }
188 for (int i = 1; i <= m; i++)
189 {
190 scanf ("%d%d", &e[i].x, &e[i].y);
191 e[i].is_del = false;
192 }
193 char op[5];
194 int tot = 0;
195 while (scanf ("%s",op) && op[0] != ‘E‘)
196 {
197 if (op[0] == ‘D‘)
198 {
199 scanf ("%d", &Q[tot].x);
200 Q[tot++].type = 0;
201 e[Q[tot-1].x].is_del = true;
202 }
203 if (op[0] == ‘Q‘)
204 {
205 scanf ("%d%d", &Q[tot].x, &Q[tot].p);
206 Q[tot++].type = 1;
207 }
208 if (op[0] == ‘C‘)
209 {
210 int v;
211 scanf ("%d%d", &Q[tot].x, &v);
212 Q[tot].p = val[Q[tot].x];
213 val[Q[tot].x] = v;
214 Q[tot++].type = 2;
215 }
216 }
217 init();
218 for (int i = 0; i < n; i++)
219 {
220 if (root[i+1] != NULL)
221 remove_tree(root[i+1]);
222 root[i+1] = new Node (val[i+1]);
223 }
224 for (int i = 0; i < m; i++)
225 {
226 if (e[i+1].is_del == false)
227 add_edge(i+1);
228 }
229 ll ans1 = 0, ans2 = 0;
230 for (int i = tot - 1; i >= 0; i--)
231 {
232 if (Q[i].type == 0)
233 {
234 add_edge(Q[i].x);
235 }
236 if (Q[i].type == 1)
237 {
238 ans1 += Get_kth(root[find(Q[i].x)], Q[i].p);
239 ans2 ++;
240 }
241 if (Q[i].type == 2)
242 {
243 int father = find(Q[i].x);
244 dele (root[father], val[Q[i].x]);
245 insert (root[father], Q[i].p);
246 val[Q[i].x] = Q[i].p;
247 }
248 }
249 printf("Case %d: %.6f\n", cas++, 1.0*ans1 / ans2);
250 }
251 return 0;
252 }
时间： 2024-11-06 01:50:51