[2015HDU Multi-School League supplementary title]hdu5299 Circles Game

Source: Internet
Author: User

Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=5299

Test instructions

There are n circles on the Euclidean plane, the circle will not intersect and will not be tangent, now Alice and Bob play the game, the two take turns to choose a circle to delete it and it contains all the circle (Alice Initiator), in their own round no circle can be deleted, ask you who will win the game

Solution: First, the relationship between the circle is abstracted, transformed into a tree, and then set the SG theorem (WoW, I don't know what is the SG theorem

To learn more, here is something related to the SG function: http://www.cnblogs.com/shjwudp/articles/4715439.html

And then there's the SG theorem (feeling like you don't want to shut it up!)

We have the following theorem: [theorem] The SG value of the leaf node is 0, and the SG value of the intermediate node is the XOR of the SG value of all its child nodes plus 1.

The problem of abstract circle is to look for, each circle directly contains the circle, read the Internet, basically is O (n^2), (there appears to be O (n logn)

Please read the code to understand it:

1 /*2 * Problem:3 * AUTHOR:SHJWUDP4 * Created TIME:2015/8/8 Saturday 15:06:485 * File name:1006.cpp6 * State:7 * Memo:8  */9#include <iostream>Ten#include <cstdio> One#include <vector> A#include <cstring> -#include <algorithm> -  the using namespacestd; -  - Const intdimension=2; -  + structPoint { -     Long LongX[dimension]; + }; A structCircle { at Point o; -     intR; - }; - structEdge { -     intu, v; -Edge (intUintv): U (U), V (v) {} in }; -  to intN; +Vector<circle>arr; -Vector<edge>edges; thevector<vector<int> >G; * BOOLcmpConstCircle & A,ConstCircle &b) { $     returnA.r>B.R;Panax Notoginseng } - voidinit () { the edges.clear (); +G.assign (n+1, vector<int> (0)); A } the voidAddedge (intUintv) { + Edges.push_back (Edge (U, v)); -G[u].push_back (Edges.size ()-1); $ } $template<classT> T Sqr (t x) {returnX *x;} - BOOLJudgeintAintb) { -     if(A==n | | b==n)return 1; the     intD=0; -      for(intk=0; k<dimension; k++) {WuyiD+=SQR (arr[a].o.x[k]-arr[b].o.x[k]); the     } -     intR=Max (ARR[A].R, ARR[B].R); Wu     returnr*r>=D; - } About voidFinintUintx) { $      for(intI:g[u]) { -Edge & e=Edges[i]; -         if(judge (x, e.v)) { - Fin (e.v, x); A             return; +         } the     } - Addedge (U, x); $ } the intDfsintu) { the     intres=0; the      for(intI:g[u]) { theEdge & e=Edges[i]; -res^=1+DFS (E.V); in     } the     returnRes; the } About intMain () { the #ifndef Online_judge theFreopen ("inch","R", stdin); the     //freopen ("Out", "w", stdout); + #endif -     intT; thescanf"%d", &T);Bayi      while(t--) { thescanf"%d", &n); the arr.resize (n); -          for(intI=0; i<n; i++) { -Circle & c=Arr[i]; the              for(intk=0; k<dimension; k++) { thescanf"%i64d", &c.o.x[k]); the             } thescanf"%d", &C.R); -         } the sort (Arr.begin (), Arr.end (), CMP); the  the init ();94          for(intI=0; i<n; i++) fin (n, i); thePuts (DFS (n)?"Alice":"Bob"); the     } the     return 0;98}
hdu5299

[2015HDU Multi-School League supplementary title]hdu5299 Circles Game

Contact Us

The content source of this page is from Internet, which doesn't represent Alibaba Cloud's opinion; products and services mentioned on that page don't have any relationship with Alibaba Cloud. If the content of the page makes you feel confusing, please write us an email, we will handle the problem within 5 days after receiving your email.

If you find any instances of plagiarism from the community, please send an email to: info-contact@alibabacloud.com and provide relevant evidence. A staff member will contact you within 5 working days.

A Free Trial That Lets You Build Big!

Start building with 50+ products and up to 12 months usage for Elastic Compute Service

  • Sales Support

    1 on 1 presale consultation

  • After-Sales Support

    24/7 Technical Support 6 Free Tickets per Quarter Faster Response

  • Alibaba Cloud offers highly flexible support services tailored to meet your exact needs.