POJ 2187 Beauty Contest(凸包求最远点距离)
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Beauty Contest
Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 30924 Accepted: 9572
Description
Bessie, Farmer John's prize cow, has just won first place in a bovine beauty contest, earning the title 'Miss Cow World'. As a result, Bessie will make a tour of N (2 <= N <= 50,000) farms around the world in order to spread goodwill between farmers and their cows. For simplicity, the world will be represented as a two-dimensional plane, where each farm is located at a pair of integer coordinates (x,y), each having a value in the range -10,000 ... 10,000. No two farms share the same pair of coordinates.
Even though Bessie travels directly in a straight line between pairs of farms, the distance between some farms can be quite large, so she wants to bring a suitcase full of hay with her so she has enough food to eat on each leg of her journey. Since Bessie refills her suitcase at every farm she visits, she wants to determine the maximum possible distance she might need to travel so she knows the size of suitcase she must bring.Help Bessie by computing the maximum distance among all pairs of farms.
Even though Bessie travels directly in a straight line between pairs of farms, the distance between some farms can be quite large, so she wants to bring a suitcase full of hay with her so she has enough food to eat on each leg of her journey. Since Bessie refills her suitcase at every farm she visits, she wants to determine the maximum possible distance she might need to travel so she knows the size of suitcase she must bring.Help Bessie by computing the maximum distance among all pairs of farms.
Input
* Line 1: A single integer, N
* Lines 2..N+1: Two space-separated integers x and y specifying coordinate of each farm
* Lines 2..N+1: Two space-separated integers x and y specifying coordinate of each farm
Output
* Line 1: A single integer that is the squared distance between the pair of farms that are farthest apart from each other.
Sample Input
40 00 11 11 0
Sample Output
2
给你一些点,求最远的两点距离的平方值
#include <iostream> //G++#include <cstdio>#include <cstring>#include <stdlib.h>#include <algorithm>#include <queue>#include <map>#include <cmath>#define inf 0x3f3f3f3f#define N 50010using namespace std;struct Point{ int x, y; int dis;} pt[N], stack[N],p0;int top, tot;int Dis(int x1 ,int y1, int x2, int y2){ return (x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2) ;}int Cmp_PolarAngel(struct Point p1, struct Point p2, struct Point pb){ int delta = (p1.x - pb.x) * (p2.y - pb.y) - (p2.x - pb.x) * (p1.y - pb.y); if(delta < 0) return 1; if(delta == 0) return 0; return -1;}bool Is_LeftTurn(struct Point p3, struct Point p2, struct Point p1){ int type = Cmp_PolarAngel(p3, p1, p2); if(type < 0) return true; return false;}int cmp(const void *p1, const void *p2){ struct Point *a1 = (struct Point*) p1; struct Point *a2 = (struct Point*) p2; int type = Cmp_PolarAngel(*a1, *a2, p0); if(type < 0) return -1; if(type == 0) { if(a1->dis < a2->dis) return -1; if(a1->dis == a2->dis) return 0; return 1; } return 1;}void Hull(int n){ int k; p0.x = inf; p0.y = inf; for(int i = 0; i < n; i++) { scanf("%d%d", &pt[i].x, &pt[i].y); if(pt[i].y < p0.y) { p0.y = pt[i].y; p0.x = pt[i].x; k = i; } else if(pt[i].y == p0.y) { if(pt[i].x < p0.x) { p0.x = pt[i].x; k = i; } } } pt[k] = pt[0]; pt[0] = p0; for(int i = 1; i < n; i++) { pt[i].dis = Dis(pt[i].x, pt[i].y, p0.x, p0.y); } qsort(pt+1, n-1, sizeof(struct Point), cmp); tot = 1; for(int i = 2; i < n; i++) { if(Cmp_PolarAngel(pt[i], pt[i-1], p0)) pt[tot++] = pt[i-1]; } pt[tot++] = pt[n-1]; top = 1; stack[0] = pt[0]; stack[1] = pt[1]; for(int i = 2; i < tot; i++) { while(top >= 1 && Is_LeftTurn(pt[i], stack[top], stack[top - 1]) == false ) { top--; } stack[++top] = pt[i]; }}int GrossProduct(struct Point p1, struct Point p2, struct Point p3){ return (p1.x - p3.x) * (p2.y - p3.y) - (p2.x - p3.x) *(p1.y - p3.y);}void Rotate(struct Point *ch, int n){ int p = 1; int t1, t2, dif; int ans = 0; ch[n] = ch[0]; for( int i = 0; i < n; i++) { while( fabs(GrossProduct(ch[i], ch[i+1], ch[p+1])) > fabs(GrossProduct(ch[i], ch[i+1], ch[p]))) { p = (p + 1) % n; } dif = fabs(GrossProduct(ch[i], ch[i+1], ch[p+1])) > fabs(GrossProduct(ch[i], ch[i+1], ch[p])); if(dif == 0) { t1 = Dis(ch[p].x, ch[p].y, ch[i].x, ch[i].y); t2 = Dis(ch[p + 1].x, ch[p + 1].y, ch[i + 1].x, ch[i + 1].y); ans = max(ans, t1); ans = max(ans, t2); } else if(dif < 0) { t1 = Dis(ch[p].x, ch[p].y, ch[i].x, ch[i].y); ans = max(ans, t1); } } printf("%d\n", ans);}int main(){ int n; while(~scanf("%d", &n)) { Hull(n); Rotate(stack , top + 1); } return 0;}
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