[ACM-ICPC] Traversal

Transcript

1. Traversal ֲࢸݢʢKuoE0ʣ KuoE0.tw@gmail.com KuoE0.ch
2. Latest update: Mar 13, 2013 Attribution-ShareAlike 3.0 Unported (CC BY-SA 3.0) http://creativecommons.org/licenses/by-sa/3.0/
3. Graph
4. Traversal
5. Traversal
6. DFS (Depth-First Search)
7. DFS (Depth-First Search)
8. DFS (Depth-First Search)
9. DFS (Depth-First Search)
10. DFS (Depth-First Search)
11. DFS (Depth-First Search)
12. DFS (Depth-First Search)
13. DFS (Depth-First Search)
14. DFS (Depth-First Search)
15. DFS (Depth-First Search)
16. DFS (Depth-First Search)
17. DFS (Depth-First Search)
18. DFS (Depth-First Search)
19. DFS (Depth-First Search)
20. DFS (Depth-First Search)
21. DFS (Depth-First Search)
22. DFS & stack 1 2 3 4 5 6
23. DFS & stack 1 2 3 4 5 6 1 1
24. DFS & stack 1 2 3 4 5 6 1 1 2 2
25. DFS & stack 1 2 3 4 5 6 1 1 2 2 3 3
26. DFS & stack 1 2 3 4 5 6 1 1 2 2 3 3 5 5
27. DFS & stack 1 2 3 4 5 6 1 1 2 2 3 3 5 5
28. DFS & stack 1 2 3 4 5 6 1 1 2 2 3 5 5 3
29. DFS & stack 1 2 3 4 5 6 1 1 2 2 3 5 5 3 4 4
30. DFS & stack 1 2 3 4 5 6 1 1 2 2 3 5 5 3 4 4 6 6
31. DFS & stack 1 2 3 4 5 6 1 1 2 2 3 5 5 3 4 4 6 6
32. DFS & stack 1 2 3 4 5 6 1 1 2 2 3 5 5 3 4 6 6 4
33. DFS & stack 1 2 3 4 5 6 1 1 2 3 5 5 3 4 6 6 4 2
34. DFS & stack 1 2 3 4 5 6 1 2 3 5 5 3 4 6 6 4 2 1
35. Source Code // There are n nodes. // g is an adjacent matrix. void DFS( int now ) { if ( visited[ now ] == true ) return; visited[ now ] = true; for ( int i = 0; i < n; ++i ) { if ( g[ now ][ i ] == true ) DFS( i ); } return; }
36. Source Code // There are n nodes. // g is an adjacent matrix. void DFS() { stack< int > stk; stk.push( 0 ); while ( stk.empty() != true ) { int now = stk.top(); visited[ now ] = true; bool noleaf = true; for ( int i = 0; i < n; ++i ) if ( g[ now ][ i ] == true && visited[ i ] != true ) { stk.push( i ); noleaf = false; break; } if ( noleaf == true ) stk.pop(); } }
37. BFS (Breadth-First Search)
38. BFS (Breadth-First Search)
39. BFS (Breadth-First Search)
40. BFS (Breadth-First Search)
41. BFS (Breadth-First Search)
42. BFS & queue 1 2 3 4 5 6 1
43. BFS & queue 1 2 3 4 5 6 1 1 1
44. BFS & queue 1 2 3 4 5 6 1 1 1
45. BFS & queue 1 2 3 4 5 6 1 1 1 2 2
46. BFS & queue 1 2 3 4 5 6 1 1 2 2
47. BFS & queue 1 2 3 4 5 6 1 1 2 2
48. BFS & queue 1 2 3 4 5 6 1 1 2 2 3 3
49. BFS & queue 1 2 3 4 5 6 1 1 2 2 3 3 4 4
50. BFS & queue 1 2 3 4 5 6 1 1 2 3 3 4 4
51. BFS & queue 1 2 3 4 5 6 1 1 2 3 3 4 4
52. BFS & queue 1 2 3 4 5 6 1 1 2 3 3 4 4 5 5
53. BFS & queue 1 2 3 4 5 6 1 1 2 3 4 4 5 5
54. BFS & queue 1 2 3 4 5 6 1 1 2 3 4 4 5 5
55. BFS & queue 1 2 3 4 5 6 1 1 2 3 4 4 5 5 6 6
56. BFS & queue 1 2 3 4 5 6 1 1 2 3 4 5 5 6 6
57. BFS & queue 1 2 3 4 5 6 1 1 2 3 4 5 5 6 6
58. BFS & queue 1 2 3 4 5 6 1 1 2 3 4 5 6 6
59. BFS & queue 1 2 3 4 5 6 1 1 2 3 4 5 6 6
60. BFS & queue 1 2 3 4 5 6 1 1 2 3 4 5 6
61. Source Code // There are n nodes. // g is an adjacent matrix. void BFS() { queue< int > que; que.push( 0 ); while ( que.empty() != true ) { int now = que.front(); que.pop(); visited[ now ] = true; for ( int i = 0; i < n; ++i ) if ( g[ now ][ i ] == true ) que.push( i ); } }
62. Time Complexity DFS O(V+E) BFS O(V+E) V: the number of nodes E: the number of edges
63. ࿝૏૸໎ٶ (DFS)
64. ࿝૏૸໎ٶ (DFS)
65. ࿝૏૸໎ٶ (BFS)
66. 1 ࿝૏૸໎ٶ (BFS)
67. 1 ࿝૏૸໎ٶ (BFS) 2
68. 1 ࿝૏૸໎ٶ (BFS) 2 3
69. 1 ࿝૏૸໎ٶ (BFS) 2 3 4
70. 1 ࿝૏૸໎ٶ (BFS) 2 3 4 5 5
71. 1 ࿝૏૸໎ٶ (BFS) 2 3 4 5 6 5 6
72. 1 ࿝૏૸໎ٶ (BFS) 2 3 4 5 6 7 5 7 6 7
73. 1 ࿝૏૸໎ٶ (BFS) 2 3 4 5 6 7 5 8 7 6 7 8
74. 1 ࿝૏૸໎ٶ (BFS) 2 3 4 5 6 7 5 8 9 7 6 7 9 9 8
75. 1 ࿝૏૸໎ٶ (BFS) 2 3 4 5 6 7 5 8 9 10 7 6 7 9 9 8 10
76. 1 ࿝૏૸໎ٶ (BFS) 2 3 4 5 6 7 11 5 8 9 10 7 6 7 9 9 8 10 11
77. 1 ࿝૏૸໎ٶ (BFS) 2 3 4 5 6 7 11 12 5 8 9 10 7 6 7 9 9 8 10 11 12
78. 1 ࿝૏૸໎ٶ (BFS) 2 3 4 5 6 7 11 12 13 5 8 9 10 7 6 7 9 9 8 10 11 12 13
79. 1 ࿝૏૸໎ٶ (BFS) 2 3 4 5 6 7 11 12 13 14 5 8 9 10 14 7 6 7 9 9 8 10 11 12 13 14
80. 1 ࿝૏૸໎ٶ (BFS) 2 3 4 5 6 7 11 12 13 14 5 8 9 10 14 7 6 7 9 9 8 10 11 12 13 14 15
81. 1 ࿝૏૸໎ٶ (BFS) 2 3 4 5 6 7 11 12 13 14 5 8 9 10 14 7 6 7 9 9 8 10 11 12 13 14 15 16
82. 1 1 ࿝૏૸໎ٶ (BFS) 2 3 4 5 6 7 11 12 13 14 5 8 9 10 14 7 6 7 9 9 8 10 11 12 13 14 15 16 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
83. 1 1 ࿝૏૸໎ٶ (BFS) 2 3 4 5 6 7 11 12 13 14 5 8 9 10 14 7 6 7 9 9 8 10 11 12 13 14 15 16 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
84. 1 1 ࿝૏૸໎ٶ (BFS) 2 3 4 5 6 7 11 12 13 14 5 8 9 10 14 7 6 7 9 9 8 10 11 12 13 14 15 16 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
85. 1 1 ࿝૏૸໎ٶ (BFS) 2 3 4 5 6 7 11 12 13 14 5 8 9 10 14 7 6 7 9 9 8 10 11 12 13 14 15 16 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
86. 1 1 ࿝૏૸໎ٶ (BFS) 2 3 4 5 6 7 11 12 13 14 5 8 9 10 14 7 6 7 9 9 8 10 11 12 13 14 15 16 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
87. 1 1 ࿝૏૸໎ٶ (BFS) 2 3 4 5 6 7 11 12 13 14 5 8 9 10 14 7 6 7 9 9 8 10 11 12 13 14 15 16 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
88. 1 1 ࿝૏૸໎ٶ (BFS) 2 3 4 5 6 7 11 12 13 14 5 8 9 10 14 7 6 7 9 9 8 10 11 12 13 14 15 16 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
89. 1 1 ࿝૏૸໎ٶ (BFS) 2 3 4 5 6 7 11 12 13 14 5 8 9 10 14 7 6 7 9 9 8 10 11 12 13 14 15 16 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
90. 1 1 ࿝૏૸໎ٶ (BFS) 2 3 4 5 6 7 11 12 13 14 5 8 9 10 14 7 6 7 9 9 8 10 11 12 13 14 15 16 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
91. 1 1 ࿝૏૸໎ٶ (BFS) 2 3 4 5 6 7 11 12 13 14 5 8 9 10 14 7 6 7 9 9 8 10 11 12 13 14 15 16 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
92. 1 1 ࿝૏૸໎ٶ (BFS) 2 3 4 5 6 7 11 12 13 14 5 8 9 10 14 7 6 7 9 9 8 10 11 12 13 14 15 16 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
93. 1 1 ࿝૏૸໎ٶ (BFS) 2 3 4 5 6 7 11 12 13 14 5 8 9 10 14 7 6 7 9 9 8 10 11 12 13 14 15 16 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
94. 1 1 ࿝૏૸໎ٶ (BFS) 2 3 4 5 6 7 11 12 13 14 5 8 9 10 14 7 6 7 9 9 8 10 11 12 13 14 15 16 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
95. 1 1 ࿝૏૸໎ٶ (BFS) 2 3 4 5 6 7 11 12 13 14 5 8 9 10 14 7 6 7 9 9 8 10 11 12 13 14 15 16 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
96. 1 1 ࿝૏૸໎ٶ (BFS) 2 3 4 5 6 7 11 12 13 14 5 8 9 10 14 7 6 7 9 9 8 10 11 12 13 14 15 16 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
97. 1 1 ࿝૏૸໎ٶ (BFS) 2 3 4 5 6 7 11 12 13 14 5 8 9 10 14 7 6 7 9 9 8 10 11 12 13 14 15 16 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
98. 1 1 ࿝૏૸໎ٶ (BFS) 2 3 4 5 6 7 11 12 13 14 5 8 9 10 14 7 6 7 9 9 8 10 11 12 13 14 15 16 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
99. 1 1 ࿝૏૸໎ٶ (BFS) 2 3 4 5 6 7 11 12 13 14 5 8 9 10 14 7 6 7 9 9 8 10 11 12 13 14 15 16 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
100. 1 1 ࿝૏૸໎ٶ (BFS) 2 3 4 5 6 7 11 12 13 14 5 8 9 10 14 7 6 7 9 9 8 10 11 12 13 14 15 16 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
101. 1 1 ࿝૏૸໎ٶ (BFS) 2 3 4 5 6 7 11 12 13 14 5 8 9 10 14 7 6 7 9 9 8 10 11 12 13 14 15 16 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
102. Practice Now [UVa] 352 - The Seasonal War
103. 352 - The Seasonal War Sample Input Sample Output 6 100100 001010 000000 110000 111000 010100 8 01100101 01000001 00011000 00000010 11000011 10100010 10000001 01100000 Image number 1 contains 3 war eagles. Image number 2 contains 6 war eagles.
104. 352 - The Seasonal War
105. 352 - The Seasonal War
106. 352 - The Seasonal War
107. 352 - The Seasonal War
108. Source Code #include <iostream> #include <cstdio> #include <cstring> using namespace std; int n; char map[ 30 ][ 30 ]; bool visited[ 30 ][ 30 ]; int step[ 8 ][ 2 ] = { { 1, 1 }, { 1, 0 }, { 1, -1 }, { 0, 1 }, { 0, -1 }, { -1, 1 }, { -1, 0 }, { -1, -1 } }; void DFS( int r, int c ); int main() { int t = 0; while ( scanf( “%d”, &n ) != EOF ) { int ret = 0; for ( int i = 0; i < n; ++i ) scanf( “%s”, map[ i ] ); memeset( visited, 0, sizeof( visited ) );
109. Source Code for ( int i = 0; i < n; ++i ) for ( int j = 0; j < n; ++j ) if ( map[ i ][ j ] == ‘1’ && visited[ i ][ j ] != true ) { DFS( i, j ); ++ret; } printf( “Image number %d contains %d war eagles.\n”, ++t, ret ); } return 0; } void DFS( int r, int c ) { if ( visited[ r ][ c ] == true ) return; visited[ r ][ c ] = true; for ( int i = 0; i < 8; ++i ) { int r2 = r + step[ i ][ 0 ], c2 = c + step[ i ][ 1 ]; if ( r2 >= 0 && r2 < n && c2 >= 0 && c2 < n && map[ r2 ][ c2 ] == ‘1’ ) DFS( r2, c2 ); } }
110. Practice Now [UVa] 260 - Il Gioco dell'X
111. Thank You for Your Listening.