[ACM-ICPC] Efficient Algorithm

Transcript

1. Efficient Algorithm ֲࢸݢʢKuoE0ʣ KuoE0.tw@gmail.com KuoE0.ch
2. Attribution-ShareAlike 3.0 Unported (CC BY-SA 3.0) http://creativecommons.org/licenses/by-sa/3.0/ Latest update: Feb 27, 2013
3. Time Complexity
4. ࠷େ࿈᠃࿨ ༗Ұ௕౓ҝ n తᏐྻ A1, A2, A3, ..., An-1, An ɼٻଖ࠷େ࿈᠃ეؒ࿨ɻ A1 A2 A3 ... An-2 An-1 An } MAX
5. int MAX = A[ 1 ]; for ( int i = 1; i <= n; ++i ) for ( int j = i; j <= n; ++j ) { int sum = 0; for ( int k = i; k <= j; ++k ) sum += A[ k ]; MAX = max( MAX, sum ); }
6. int MAX = A[ 1 ]; for ( int i = 1; i <= n; ++i ) for ( int j = i; j <= n; ++j ) { int sum = 0; for ( int k = i; k <= j; ++k ) sum += A[ k ]; MAX = max( MAX, sum ); } ֩৺ӡࢉɿՃ๏
7. int MAX = A[ 1 ]; for ( int i = 1; i <= n; ++i ) for ( int j = i; j <= n; ++j ) { int sum = 0; for ( int k = i; k <= j; ++k ) sum += A[ k ]; MAX = max( MAX, sum ); }
8. int MAX = A[ 1 ]; for ( int i = 1; i <= n; ++i ) for ( int j = i; j <= n; ++j ) { int sum = 0; for ( int k = i; k <= j; ++k ) sum += A[ k ]; MAX = max( MAX, sum ); } T(n) = j − i +1 j=i n ∑ i=1 n ∑ = n(n +1)(n + 2) 6 Ճ๏࣍Ꮠɿ
9. T(n) = j − i +1 j=i n ∑ i=1 n ∑ = n(n +1)(n + 2) 6
10. T(n) = j − i +1 j=i n ∑ i=1 n ∑ = n(n +1)(n + 2) 6 ࡾ࣍ࣜ
11. T(n) = j − i +1 j=i n ∑ i=1 n ∑ = n(n +1)(n + 2) 6 ࡾ࣍ࣜ n 3 n 2 n 1 n=10 10 3 10 2 10 1 1 n=10 2 10 6 10 4 10 2 1 n=10 3 10 9 10 6 10 3 1
12. 0 1750 3500 5250 7000 3 7 11 13 17 19 n^3 n^2 n 1
13. ࠷େࢦᏐ߲࣍Өڹ࠷େ Time Complexity: O(n3)
14. ෆधਫ਼֬෼ੳ ਐߦਫ਼֬෼ੳֱ໣࣌ؒʂ
15. ෆधਫ਼֬෼ੳ ུဟ᫮ᅲ࣍Ꮠ ਐߦਫ਼֬෼ੳֱ໣࣌ؒʂ
16. int MAX = A[ 1 ]; for ( int i = 1; i <= n; ++i ) for ( int j = i; j <= n; ++j ) { int sum = 0; for ( int k = i; k <= j; ++k ) sum += A[ k ]; MAX = max( MAX, sum ); } ࠷ଟ࣍Ꮠ ୈ 1 ૚᫮ᅲ n ୈ 2 ૚᫮ᅲ n ୈ 3 ૚᫮ᅲ n
17. n × n × n = n3 Time Complexity: O(n3)
18. લ႔ཧ༏Խ
19. ༗Ұ௕౓ҝ n తᏐྻ A1, A2, A3, ..., An-1, An ɼٻଖ࠷େ࿈᠃ეؒ࿨ɻ } MAX ࠷େ࿈᠃࿨ A1 A2 A3 ... An-2 An-1 An
20. ઃᏐྻ S ҝᏐྻ A తྦྷੵ࿨ɼ Ҽࠑ Si = A1 + A2 + ... + Ai ɼଇ Ai + Ai+1 + ... + Aj = Sj - Si-1 ɻ
21. ઃᏐྻ S ҝᏐྻ A తྦྷੵ࿨ɼ Ҽࠑ Si = A1 + A2 + ... + Ai ɼଇ Ai + Ai+1 + ... + Aj = Sj - Si-1 ɻ ٻეؒ [i,j] త࿈᠃࿨ᷮधཁ O(1) త࣌ؒɻ
22. A1 A2 A3 A4 A5 A6 A7 A8 S1 3 7 -1 -3 9 -10 1 2 3 S2 3 7 -1 -3 9 -10 1 2 10 S3 3 7 -1 -3 9 -10 1 2 9 S4 3 7 -1 -3 9 -10 1 2 6 S5 3 7 -1 -3 9 -10 1 2 15 S6 3 7 -1 -3 9 -10 1 2 5 S7 3 7 -1 -3 9 -10 1 2 6 S8 3 7 -1 -3 9 -10 1 2 8
23. int MAX = A[ 1 ]; S[ 0 ] = 0; for ( int i = 1; i <= n; ++i ) S[ i ] = S[ i - 1 ] + A[ i ]; for ( int i = 1; i <= n; ++i ) for ( int j = i; j <= n; ++j ) { MAX = max( MAX, S[ j ] - S[ i - 1 ] ); }
24. int MAX = A[ 1 ]; S[ 0 ] = 0; for ( int i = 1; i <= n; ++i ) S[ i ] = S[ i - 1 ] + A[ i ]; for ( int i = 1; i <= n; ++i ) for ( int j = i; j <= n; ++j ) { MAX = max( MAX, S[ j ] - S[ i - 1 ] ); } O(n) O(n2) ࠷େࢦᏐ߲࣍ɿO(n2)
25. Divide & Conquer ෼࣏ࣕ೭
26. ֢෼ҝ૬ಉతࢠ໰୊
27. ֢෼ҝ૬ಉతࢠ໰୊ ᬇ᫮ٻղࢠ໰୊
28. ֢෼ҝ૬ಉతࢠ໰୊ ᬇ᫮ٻղࢠ໰୊ ሡࢠ໰୊తղ߹ซ
29. ࠷େ࿈᠃࿨ ༗Ұ௕౓ҝ n తᏐྻ A1, A2, A3, ..., An-1, An ɼٻଖ࠷େ࿈᠃ეؒ࿨ɻ } MAX A1 A2 A3 ... An-2 An-1 An
30. ੾ׂ໰୊ A1 A2 A3 ... An-2 An-1 An
31. ੾ׂ໰୊ A1 A2 ... An/2 A1 A2 A3 ... An-2 An-1 An An/2+1 An/2+2 ... An An/4+1 ... An/2 A1 ... An/4 An/2+1 ... A3n/4 A3n/4+1 ... An A1 An/4 An/4+1 An/2 An/2+1 A3n/4 A3n/4+1 An
32. ٻղ໰୊ ᬇ᫮௚౸ᏐྻᷮႫҰݩૉɼ௚઀ճၚ֘Ꮠ值ɻ ᷮ༗Ұݩૉ࣌ɼ֘Ꮠ值ҝ࠷େ࿈᠃࿨ɻ
33. ߹ซࢠ໰୊ղ 10 11 બᎩࢠ໰୊த࠷༏ղʂ ... ...
34. ߹ซࢠ໰୊ղ 10 11 બᎩࢠ໰୊த࠷༏ղʂ ... ...
35. But.... A1 A2 ... An/2 A1 A2 A3 ... An-2 An-1 An An/2+1 An/2+2 ... An
36. But.... ౴ҊՄೳℷލၷݸეؒʂ A1 A2 ... An/2 A1 A2 A3 ... An-2 An-1 An An/2+1 An/2+2 ... An
37. How-to 5 -9 3 1 3 -10 3 3
38. How-to 5 -9 3 1 3 -10 3 3 5 ࠷େ࿈᠃࿨ɿ5
39. How-to 5 -9 3 1 3 -10 3 3 3 5 3 ࠷େ࿈᠃࿨ɿ5 ࠷େ࿈᠃࿨ɿ6
40. How-to 5 -9 3 1 3 -10 3 3 3 5 3 ࠷େ࿈᠃࿨ɿ5 ࠷େ࿈᠃࿨ɿ6 ᙛલ࠷େ࿈᠃࿨ɿ6
41. ᙛલ࠷େ࿈᠃࿨ɿ6 How-to 5 -9 3 1 3 -10 1 1
42. ᙛલ࠷େ࿈᠃࿨ɿ6 How-to 5 -9 3 1 3 -10 1 1 ࣗӈࣕࠨፙ࠷େ࿈᠃࿨ 1 3
43. ᙛલ࠷େ࿈᠃࿨ɿ6 How-to 5 -9 3 1 3 -10 1 1 ࣗӈࣕࠨፙ࠷େ࿈᠃࿨ 1 3 ࿈᠃࿨ɿ4
44. ᙛલ࠷େ࿈᠃࿨ɿ6 How-to 5 -9 3 1 3 -10 1 1 ࣗӈࣕࠨፙ࠷େ࿈᠃࿨ ࣗࠨࣕӈፙ࠷େ࿈᠃࿨ 1 3 3 ࿈᠃࿨ɿ4
45. ᙛલ࠷େ࿈᠃࿨ɿ6 How-to 5 -9 3 1 3 -10 1 1 ࣗӈࣕࠨፙ࠷େ࿈᠃࿨ ࣗࠨࣕӈፙ࠷େ࿈᠃࿨ 1 3 3 ࿈᠃࿨ɿ4 ࿈᠃࿨ɿ3
46. ᙛલ࠷େ࿈᠃࿨ɿ6 How-to 5 -9 3 1 3 -10 1 1 ࣗӈࣕࠨፙ࠷େ࿈᠃࿨ ࣗࠨࣕӈፙ࠷େ࿈᠃࿨ 1 3 3 ࿈᠃࿨ɿ4 ࿈᠃࿨ɿ3 ࿈᠃࿨ɿ7
47. ࠷ऴ࠷େ࿈᠃࿨ɿ7 ᙛલ࠷େ࿈᠃࿨ɿ6 How-to 5 -9 3 1 3 -10 1 1 ࣗӈࣕࠨፙ࠷େ࿈᠃࿨ ࣗࠨࣕӈፙ࠷େ࿈᠃࿨ 1 3 3 ࿈᠃࿨ɿ4 ࿈᠃࿨ɿ3 ࿈᠃࿨ɿ7
48. int maxsum( int L, int R ) { if ( R - L == 1 ) return A[ L ]; int mid = ( L + R ) / 2; int MAX = max( maxsum( L, mid ), maxsum( mid, R ) ); int tempL = 0, tempR = 0; for ( int i = mid - 1, sum = 0; i >= L; --i ) tempL = max( tempL, sum += A[ i ] ); for ( int i = mid, sum = 0; i < R; ++i ) tempR = max( tempR, sum += A[ i ] ); return max( MAX, tempL + tempR ); }
49. Time Complexity total time
50. Time Complexity total time n 2 * (n/2) = n ... n * (n/n) = n
51. The height }log2n
52. Time Complexity: O(n×log2n)
53. Time Complexity of Divide & Conquer ࢠ໰୊ፒআᬇ᫮ݺڣత࣌ؒෳᯑ౓ C(n) ᬇ᫮ݺڣ૚Ꮠ H(n) Time Complexity: C(n) × H(n)
54. Time Comparison n3 n2 nlog2n 100 1000 0.15 0.02 0.01 123.86 1.36 0.14 ᄸҐɿඵ
55. ኺଌࢼࢿྉ൑Ꮧԋࢉ๏ ၊ઃػث 1 ඵՄҎࣥߦ 106 ݸࢦྩɼ... ࣌ؒෳᯑ౓ n! 2n n3 n2 nlog2n n ࠷େن໛ 9 20 10 2 10 3 7.5 × 10 4 10 6
56. 㠥Ұछ㘸๏ ၊ઃػث 1 ඵՄҎࣥߦ 108 ݸࢦྩɼ... ࣌ؒෳᯑ౓ n! 2n n3 n2 nlog2n n ࠷େن໛ 11 26 464 10 4 4.5 × 10 6 10 8
57. 10245 - The Closest Pair Problem Practice Now
58. Reference •ఏঋఔࣜઃܭతᬓाࢥߟྗ (link)
59. Thank You for Your Listening.