Codeforces 496D Tennis Game enumeration + two points

Given n games.

The following n numbers: Indicates whether the field is 1 wins or 2 wins.

1. The winner gets a point.

2, if the whole season has been decided the outcome of the game will not continue.

3, set to win this season need to win the S-Bureau, each innings first won the T-score player.

Ask:

Find all the (s,t) combinations to make the given N game record legal.

The output to sort.

Enumerates T.

A array of what matches the first person wins.

b array the number of matches that the second person won.

Set to win T is divided into one sentence. Then the first person to win the T-score is the second, and the next one wins the T-Score is the third.

The man who won the T-Score obviously won the game.

This runs at the same time a number of 2 people.

Complexity of:

Set to win a score of I, then this array is enumerated up to n/i times

i = [1, n]

So complexity = N/1 + N/2 + + n/n = NLN (n+1);

There is a two-point. The total complexity is O (n * LN (n+1) * log n);

[CPP]View Plaincopy

2. Import java.util.ArrayList;
4. Import java.util.Collections;
6. Import java.util.List;
8. Import Java.util.Scanner;
10. Public class Main {
12. class Ans implements comparable<ans>{
14. int S, t;
16. Public Ans (int A, int b) {
18. s = A; t = b;
20. }
22. public int compareTo (Ans o) {
24. if (s! = O.s)return Integer.compare (S, O.S);
26. return Integer.compare (T, O.T);
28. }
30. }
32. static int N = 100050;
34. int[] A = new Int[n], B = new int[n], w = new int[n];
36. int at, BT, N;
38. arraylist<ans> Ans = new arraylist<> ();
40. void init () {
42. n = cin.nextint ();
44. For (int i = 1; I <= n; i++)
46. W[i] = Cin.nextint ();
48. at = BT = 0;
50. a[at++] = 0; b[bt++] = 0;
52. For (int i = 1; I <= n; i++)
54. {
56. if (w[i]==1)
58. a[at++] = i;
60. Else
62. b[bt++] = i;
64. }
66. at--; bt--;
68. Ans.clear ();
70. }
72. void Add (int x, int y) {
74. Ans tmp = new ans (x, y);
76. Ans.add (TMP);
78. }
80. void Work (int x) {
82. int na = 0, nb = 0;
84. int siza = 0, Sizb = 0;
86. //System.out.println (x+ ":" +at+ "" +BT ");
88. While (true) {
90. if (na = = at && nb = = BT) break;
92. if (na+x > at && nb+x > BT)return;
94. /* if (Na+x > at)
96. {
98. if (nb+x = = BT) {
100. sizb++; break;
102. }
104. return;
106. }
108. if (Nb+x > BT)
110. {
112. if (na+x = = at) {
114. siza++; break;
116. }
118. return;
120. }/**/
122. if (na+x<=at && (NB+X>BT | | a[na+x] < B[NB+X]))
124. {
126. siza++;
128. Na + = x;
130. int L = 0, R = bt, pos = 0;
132. While (L <= R)
134. {
136. int mid = (l+r) >>1;
138. if (B[mid] < A[na])
140. {
142. pos = mid;
144. L = mid+1;
146. }
148. Else
150. R = mid-1;
152. }
154. NB = pos;
156. }
158. Else if (nb+x <= bt)
160. {
162. sizb++;
164. NB + = x;
166. int L = 0, R = at, pos = 0;
168. While (L <= R)
170. {
172. int mid = (l+r) >>1;
174. if (A[mid] < B[NB])
176. {
178. pos = mid;
180. L = mid+1;
182. }
184. Else
186. R = mid-1;
188. }
190. na = pos;
192. }
194. else return;
196. //System.out.println ("[" +na+ "" +nb+ "]" + "{" +siza+ "" +sizb+ "}");
198. }/**/  
200. //System.out.println ("Last:" +siza+ "" +sizb);
202. if (Siza = = Sizb)return;
204. if (Siza>sizb && w[n] = = 1)
206. Add (Siza, x);
208. Else if (Siza<sizb && w[n] = = 2)
210. Add (Sizb, x);
212. }
214. public void work () {
216. Init ();
218. //int AAA = 0; if (aaa<=0) {System.out.println ("Debug"); return;}
220. For (int i = 1; I <= n; i++)
222. Work (i);
224. System.out.println (Ans.size ());
226. Collections.sort (ANS);
228. For (int i = 0; i < ans.size (); i++)
230. System.out.println (Ans.get (i). S+"" +ans.get (i). T);
232. }
234. Main () {
236. cin = New Scanner (system.in);
238. }
240. public static void Main (string[] args) {
242. Main E = new Main ();
244. E.work ();
246. }
248. Public Scanner Cin;
250. }

Codeforces 496D Tennis Game enumeration + two points