1 / 24

240 likes | 403 Views

ACM reminders. October 30 -- HMC Mock contest. Good times ?. November 10 -- ACM contest. current list of competitors. Accepted Harvey Mudd College : HMC 1 Accepted Harvey Mudd College : HMC 2 Accepted Harvey Mudd College : HMC 3 Pending Mount San Antonio College : MtSAC@Voidies$qf

Download Presentation
## ACM reminders

**An Image/Link below is provided (as is) to download presentation**
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.
Content is provided to you AS IS for your information and personal use only.
Download presentation by click this link.
While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

**ACM reminders**October 30 -- HMC Mock contest Good times ? November 10 -- ACM contest current list of competitors Accepted Harvey Mudd College: HMC 1 Accepted Harvey Mudd College: HMC 2 Accepted Harvey Mudd College: HMC 3 Pending Mount San Antonio College: MtSAC@Voidies$qf Pending Mount San Antonio College: MtSAC@Floaties$qi Pending Riverside Community College: Platinum Pending Riverside Community College: The Code Machine Pending Riverside Community College: CodeBlue Pending University of California, San Diego: UCSD Rock Pending University of California, San Diego: UCSD Paper Pending University of California, San Diego: UCSD Scissors Pending University of Southern California: Trojan Horse Pending University of Southern California: Trojan Pride Pending University of Southern California: Trojans Pending Azusa Pacific University: 4Alpha2Omega Pending Brigham Young University -- Hawaii Campus: C-Siders 1 Pending Brigham Young University -- Hawaii Campus: C-Siders 2 Pending California Lutheran University: Java the Hut Pending California Lutheran University: Just Wanna Program Pending California Lutheran University: javac this.java Pending California State University Long Beach: Beach1 Pending California State University Long Beach: Beach2 Pending California State University Long Beach: Beach3 Pending California State University, Northridge: CSUN Red Pending California State University, Northridge: CSUN Black Pending California State University, Northridge: CSUN-3 Pending El Camino College: ECC Beta November 13 -- Final acm class & contest wrap-up**C++ Map**www.dinkumware.com/htm_cpl/index.html www.sgi.com/tech/stl/ map multimap #include <map> #include <map> set of key/value pairs**C++ Map**www.dinkumware.com/htm_cpl/index.html www.sgi.com/tech/stl/ map multimap #include <map> #include <map> set of key/value pairs (implemented as a balanced binary tree)**C++ Map**www.dinkumware.com/htm_cpl/index.html www.sgi.com/tech/stl/ map multimap #include <map> #include <map> set of key/value pairs Fast: searching by key Slow: searching by value (implemented as a balanced binary tree)**C++ Map**#define MP make_pair typedef pair<int,int> PII; map<PII,int> m; m[MP(0,1)] = 10; m[MP(0,2)] = 17; m[MP(1,2)] = 5; m[MP(2,0)] = 12; // probably not worth it for graphs... map<string,int> m; // definitely worth it here m[“ArcOS”] = 110; // as an associative array m[“TheoComp”] = 140; multimap<string,string> d; // as a dictionary d.insert(MP(“fun”,“c++ coding”)); // methods exist to get d.insert(MP(“fun”,“ACM coding”)); // all of “fun”s entries 10 0 1 17 5 12 2**Geometric Problems**Problem 1 - Binary Space Partitions Basic idea: draw objects from far (first) to near (last). observer**Geometric Problems**Problem 1 - Binary Space Partitions z (50,240) (20,240) C (-30,230) (-60,230) (100,220) (70,220) B (50,220) (20,220) D (-30,210) (-60,210) (100,200) (70,200) (-20,80) (-50,80) A (70,70) (40,70) E (-20,60) (-50,60) (70,50) (40,50) x Basic idea: draw objects from far (first) to near (last). (0,-big) observer**Geometric Problems**Problem 1 - Binary Space Partitions Input: 5 4 -50 60 -20 60 -20 80 -50 80 4 -60 210 -30 210 -30 230 -60 230 4 20 220 50 220 50 240 20 240 4 70 200 100 200 100 220 70 220 4 40 50 70 50 70 70 40 70 first part**Geometric Problems**Problem 1 - Binary Space Partitions z (50,240) (20,240) C (-30,230) (-60,230) (100,220) (70,220) B (50,220) (20,220) D (-30,210) (-60,210) (100,200) (70,200) (0,140) (-20,80) (-50,80) A (70,70) (40,70) E (-20,60) (-50,60) (70,50) (40,50) (-15,0) x Basic idea: draw objects from far (first) to near (last). (0,-big) observer**Geometric Problems**Problem 1 - Binary Space Partitions z C B D (0,140) A E CDE AB (-15,0) x Basic idea: draw objects from far (first -- LEFT) to near (last -- RIGHT). (0,-big) observer**Geometric Problems**Problem 1 - Binary Space Partitions z C B D (-70,150) (70,150) (0,140) A E CDE AB E CD A B (-15,0) x Basic idea: draw objects from far (first -- LEFT) to near (last -- RIGHT). (0,-big) observer**Geometric Problems**Problem 1 - Binary Space Partitions Input: 5 4 -50 60 -20 60 -20 80 -50 80 4 -60 210 -30 210 -30 230 -60 230 4 20 220 50 220 50 240 20 240 4 70 200 100 200 100 220 70 220 4 40 50 70 50 70 70 40 70 2 0 140 -15 0 70 150 -70 150 first part second part**Geometric Problems**Problem 1 - Binary Space Partitions Input: 5 4 -50 60 -20 60 -20 80 -50 80 4 -60 210 -30 210 -30 230 -60 230 4 20 220 50 220 50 240 20 240 4 70 200 100 200 100 220 70 220 4 40 50 70 50 70 70 40 70 2 0 140 -15 0 70 150 -70 150 first part second part the objects, in the order they would be rendered by this BSP Output: ECDAB**Geometric Problems**Problem 2 - Visualizing cubes 3 3 1 3 1 2**Geometric Problems**Problem 2 - Visualizing cubes 3 3 1 3 1 2 left wall right wall floor left wall floor Suppose you rotate so that floor right wall right wall left wall What is the resulting stacking pattern?**Geometric Problems**Problem 2 - Visualizing cubes 3 3 1 3 1 2 left wall right wall floor left wall floor Suppose you rotate so that floor right wall 3 2 1 2 1 1 2 1 right wall left wall What is the resulting stacking pattern?**Geometric Problems**Problem 2 - Visualizing cubes ?!?**0**8 13 - 1 - 0 - 6 12 - 9 0 - - 7 - 0 0 - - - - 11 0 All-pairs shortest paths “Floyd-Warshall algorithm” B 8 9 to A D E B C A 13 C A 6 B 0 from C 7 1 12 D D E E 11**0**8 13 - 1 - 0 - 6 12 - 9 0 - - 7 - 0 0 - - - - 11 0 All-pairs shortest paths “Floyd-Warshall algorithm” B 8 9 to A D E B C A 13 C A 6 B 0 from C 7 1 12 D D E E 11 0 D0 = (dij ) 0 dij = shortest distance from i to j through no nodes k dij = shortest distance from i to j through nodes {1, …, k}**0**8 13 - 1 - 0 - 6 12 - 9 0 - - 7 - 0 0 - - - - 11 0 All-pairs shortest paths “Floyd-Warshall algorithm” B 8 9 to A D E B C A 13 C A 6 B 0 from C 7 1 12 D D E E 11 0 D0 = (dij ) 1 dij = shortest distance from i to j through nodes {1} k dij =**0**0 8 8 13 13 - - 1 1 - - 0 0 - - 6 6 12 12 - - 9 9 0 0 - - - - 7 7 - - 0 0 0 0 - - - - - - - - 11 11 0 0 All-pairs shortest paths... “Floyd-Warshall algorithm” A k B dij = C D E 0 D0 = (dij ) A A 0 8 13 - 1 B B - 0 - 6 12 C C - 9 0 - - D D 7 15 0 0 8 E E - - - 11 0 1 D1 = (dij ) k dij = shortest distance from i to j through nodes {1, …, k}**0**0 8 8 13 13 - - 1 1 - - 0 0 - - 6 6 12 12 - - 9 9 0 0 - - - - 7 7 - - 0 0 0 0 - - - - - - - - 11 11 0 0 All-pairs shortest paths... “Floyd-Warshall algorithm” A k B dij = C D E 0 D0 = (dij ) A A 0 8 13 - 1 B B - 0 - 6 12 C C - 9 0 - - D D 7 15 0 0 8 E E - - - 11 0 1 D1 = (dij ) k dij = shortest distance from i to j through {1, …, k}**0**0 8 8 13 13 - - 1 1 - - 0 0 - - 6 6 12 12 - - 9 9 0 0 - - - - 7 7 - - 0 0 0 0 - - - - - - - - 11 11 0 0 All-pairs shortest paths... “Floyd-Warshall algorithm” A k B dij = C D E 0 D0 = (dij ) A A 0 8 13 - 1 B B - 0 - 6 12 C C - 9 0 - - D D 7 15 0 0 8 E E - - - 11 0 1 D1 = (dij ) k dij = shortest distance from i to j through {1, …, k}**All-pairs shortest paths...**A 0 8 13 14 1 A 0 8 13 14 1 B 13 0 6 6 12 B - 0 - 6 12 C 22 9 0 15 21 C - 9 0 15 21 D 7 9 0 0 8 D 7 15 0 0 8 E 18 20 11 11 0 E - - - 11 0 4 D4 = (dij ) 2 D2 = (dij ) A 0 8 12 12 1 A 0 8 13 14 1 B 13 0 6 6 12 B - 0 - 6 12 C 22 9 0 15 21 C - 9 0 15 21 D 7 9 0 0 8 D 7 9 0 0 8 E 18 20 11 11 0 E - - - 11 0 5 3 D5 = (dij ) D3 = (dij ) to store the path, another matrix can track the last intermediate vertex

More Related