Ultraviolet A Problem Solution: 10089-repackaging

 

First, convert each package P = (S1, S2, S3) to a vector (S2-S1, S3-S1 ). all the packages forms convex hull. if point (0, 0) is in this convex hull, then there will be a solution, otherwise not. I can not tell the proof of this now, if you have any ideas, please let me know.

 

In fact, point (0, 0) is not in this convex hull if and only if all the vectors fall into a sector whose angle is less than Pi. this is obviusly correct. so we can use this point to get our answer.

 

BTW, you must care about data overflow. I 've got AC after lots of was just by change date type from int to long.

 

/*************************************** ********************************** <Br/> * copyright (c) 2008 by Liu kaipeng * <br/> * liukaipeng at gmail dot com * <br/> ******************** **************************************** * **************/<br/>/* @ judge_id 00000 10089 C ++ "repackaging" */<br/> # include <algorithm> <br/> # include <cstdio> <br/> # include <cstring> <br/> # include <deque> <br/> # include <Fstream> <br/> # include <iostream> <br/> # include <list> <br/> # include <map> <br/> # include <queue> <br/> # include <set> <br/> # include <stack> <br/> # include <string> <br/> # include <vector> <br/> using namespace STD; </P> <p> typedef long llint; <br/> llint const maxn = 1001; <br/> bool repackage (llint pkgs [] [3], llint npkgs) <br/> {<br/> for (llint I = 0; I <npkgs; ++ I) {<br/> pkgs [I] [0]-= pkgs [I] [2]; <br/> pkgs [I] [1]-= pkgs [I] [2]; <br/>}</P> <p> llint X1 = pkgs [0] [0], Y1 = pkgs [0] [1], x2 = pkgs [0] [0], y2 = pkgs [0] [1]; <br/> for (llint I = 1; I <npkgs; ++ I) {<br/> llint x = pkgs [I] [0], y = pkgs [I] [1]; <br/> llint a1 = Y1 * X, b1 = x1 * y, a2 = Y2 * X, b2 = x2 * Y; <br/> If (a1 = b1) <br/> If (x1 * x + Y1 * Y <= 0) return true; <br/> else continue; <br/> If (A2 = b2) <br/> If (X2 * x + Y2 * Y <= 0) return true; <br/> else continue; <br/> If (A1 <B1) x1 = x, Y1 = y; <br/> If (A2> B2) x2 = x, y2 = y; <br/> If (x1 = X2 & Y1 = Y2) return true; <br/> If (Y1 * X2 = Y2 * X1) <br/> If (x1 * X2 + Y1 * Y2 <= 0) return true; <br/> else continue; <br/> If (Y1 * X2 <Y2 * X1) return true; <br/>}< br/> return false; <br/>}</P> <p> int main (INT argc, char * argv []) <br/>{< br/> # ifndef onl Ine_judge <br/> freopen (string (argv [0]) + ". in "). c_str (), "r", stdin); <br/> freopen (string (argv [0]) + ". out "). c_str (), "W", stdout); <br/> # endif <br/> llint pkgs [maxn] [3]; <br/> for (llint npkgs; cin> npkgs & npkgs! = 0;) {<br/> for (llint I = 0; I <npkgs; ++ I) <br/> CIN> pkgs [I] [0]> pkgs [I] [1]> pkgs [I] [2]; <br/> If (repackage (pkgs, npkgs) <br/> cout <"Yes/N "; <br/> else <br/> cout <"No/N"; <br/>}< br/> return 0; <br/>}< br/>