稳定凸包问题 要求每条边上至少有三个点,且对凸包上点数为1,2时要特判
巨坑无比,调了很长时间= =
//POJ 1228
//稳定凸包问题,等价于每条边上至少有三个点,但对m = 1(点)和m = 2(线段)要特判
//AC 2016-10-15
#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <cmath>
#define MAXN 1010
double sqr(double x){
return x * x;
}
struct point{
int x, y;
point(){}
point(int X, int Y): x(X), y(Y){}
friend int operator ^ (const point &p1, const point &p2){
return p1.x *p2.y - p1.y * p2.x;
}
friend int operator * (const point &p1, const point &p2){
return p1.x *p2.x + p1.y * p2.y;
}
double norm(){
return sqrt(sqr(x) + sqr(y));
}
friend point operator >> (const point &p1, const point &p2){
return point(p2.x - p1.x, p2.y - p1.y);
}
friend bool operator < (const point &p1, const point &p2){
return (p1.x < p2.x)||(p1.x == p2.x)&&(p1.y < p2.y);
}
}pt[MAXN];
template <typename T>
void swap(T &a, T &b){
T t = a;
a = b;
b = t;
}
template <typename T>
void BBS(T a[], int n){
for (int i = 0; i < n; i++)
for (int j = 0; j < i; j++)
if (a[i] < a[j]) swap(a[i], a[j]);
}
bool stable_convex_hull(point p[], int n){
int res = 0, cur = 0, m = 0;
BBS(p, n);
while(1){
int tmp = - 1;
bool stable = 0;
for (int i = 0; i < n; i++)
if (i != cur)
if (!(tmp + 1)){
tmp = i, stable = 0;
}
else{
int det = (p[cur] >> p[i]) ^ (p[cur] >> p[tmp]);
if (det > 0){
tmp = i, stable = 0;
}
else if ((!det)&&((p[cur] >> p[i]) * (p[cur] >> p[tmp]) > 0)){
if ((p[cur] >> p[i]).norm() > (p[cur] >> p[tmp]).norm())
tmp = i;
stable = 1;
}
}
if (tmp + 1){
m++;
if (!stable)
return 0;
}
if (!tmp||!(tmp + 1)) return ((tmp + 1) && (m > 2));
cur = tmp;
}
}
int main(){
int t, n;
freopen("fin.c", "r", stdin);
scanf("%d", &t);
while(t--){
scanf("%d", &n);
for (int i = 0; i < n; i++){
scanf("%d%d", &pt[i].x, &pt[i].y);
}
if (stable_convex_hull(pt, n)){
puts("YES");
}
else puts("NO");
}
}
POJ 1228 - Grandpa's Estate 稳定凸包
原文:http://www.cnblogs.com/xlnx/p/5964344.html