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[andmenj-acm.git] / UVa / 10167 - Birthday cake / 10167.3.cpp
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1 /*
2 Problem: 10167 - Birthday cake
3 (From the UVa Online Judge)
4 http://acm.uva.es/problemset/v101/10167.html
6 Divide-and-conquer solution
7 (Read problem "10077 - The Stern-Brocot number system" for an idea about the divide-and-conquer argument)
8 Basically I binary-search the perfect slope to cut the cake. It really isn't BINARY search, but every
9 time the space of slopes that can possibly be a solution is reduced considerably.
11 Author: Andrés Mejía-Posada
12 Date: May 6, 2008
14 This code gets Accepted in the online judge.
17 #include <iostream>
18 #include <vector>
19 #include <algorithm>
21 using namespace std;
23 int n;
25 struct point{
26 int x, y;
27 point() {}
28 point(const int &X, const int &Y) : x(X), y(Y) {}
31 struct line{
32 int a, b;
33 line() {}
34 line(const int &A, const int &B) : a(A), b(B) {}
37 line operator |(const line &a, const line &b){
38 line r;
39 r.a = a.a + b.a;
40 r.b = a.b + b.b;
41 return r;
44 //Returns positive if p is to the left of line l
45 //Returns negative if p is to the right of line l
46 //Returns zero if p is colineal with line l
47 int cross(const point &p, const line &l){
48 point q;
49 q.x = l.b;
50 q.y = -l.a;
52 return (p.x*q.y - q.x*p.y);
55 //Returns positive if line l should rotate left
56 //Returns negative if line l should rotate right
57 //Returns zero if line l is a solution
58 int check(const vector<point> &c, const line &l){
59 int left = 0, right = 0;
60 for (int i=0; i<2*n; ++i){
61 int t = cross(c[i], l);
62 left += (t>0);
63 right += (t<0);
65 if (left == right && left < n){ //2 or more points are colineal with the cut
66 return -1; //? Or return 1? Does it matter? (It also gets accepted with +1, 0 will give a wrong answer though).
68 return (left - right);
72 int main(){
73 while (scanf("%d", &n) && n){
74 vector<point> cherry(2*n);
75 for (int i=0; i<2*n; ++i){
76 scanf("%d %d", &cherry[i].x, &cherry[i].y);
79 line uphigh, upmed, updown, downhigh, downmed, downdown;
80 uphigh = line(1, 0);
81 if (check(cherry, uphigh) == 0){
82 printf("%d %d\n", uphigh.a, uphigh.b);
83 continue;
85 updown = line(0, 1);
86 if (check(cherry, updown) == 0){
87 printf("%d %d\n", updown.a, updown.b);
88 continue;
91 downhigh = line(0, 1);
92 if (check(cherry, downhigh) == 0){
93 printf("%d %d\n", downhigh.a, downhigh.b);
94 continue;
97 downdown = line(-1, 0);
98 if (check(cherry, downdown) == 0){
99 printf("%d %d\n", downdown.a, downdown.b);
100 continue;
103 while (true){
106 upmed = uphigh | updown;
107 int t = check(cherry, upmed);
108 if (t == 0){
109 printf("%d %d\n", upmed.a, upmed.b);
110 break;
111 }else if (t < 0){
112 uphigh = upmed;
113 }else{
114 updown = upmed;
117 downmed = downhigh | downdown;
118 t = check(cherry, downmed);
119 if (t == 0){
120 printf("%d %d\n", downmed.a, downmed.b);
121 break;
122 }else if (t < 0){
123 downhigh = downmed;
124 }else{
125 downdown = downmed;
131 return 0;