【POJ】2653 Pick-up sticks（计算几何+暴力）

http://poj.org/problem?id=2653

我很好奇为什么这样$O(n^2)$的暴力能过....

虽然说这是加了链表优化的，但是最坏不也是$O(n^2)$吗。。。（只能说数据太弱...）

然后本题裸的判线段相交和点在直线上...（看了网上的标程，不判端点的情况都能过我也是醉了...）

#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <algorithm>
#include <queue>
#include <set>
#include <map>
using namespace std;
typedef long long ll;
#define pii pair<int, int>
#define mkpii make_pair<int, int>
#define pdi pair<double, int>
#define mkpdi make_pair<double, int>
#define pli pair<ll, int>
#define mkpli make_pair<ll, int>
#define rep(i, n) for(int i=0; i<(n); ++i)
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);--i)
#define for4(i,a,n) for(int i=(a);i>(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define read(a) a=getint()
#define print(a) printf("%d", a)
#define dbg(x) cout << (#x) << " = " << (x) << endl
#define error(x) (!(x)?puts("error"):0)
#define printarr2(a, b, c) for1(_, 1, b) { for1(__, 1, c) cout << a[_][__]; cout << endl; }
#define printarr1(a, b) for1(_, 1, b) cout << a[_] << ‘\t‘; cout << endl
inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<‘0‘||c>‘9‘; c=getchar()) if(c==‘-‘) k=-1; for(; c>=‘0‘&&c<=‘9‘; c=getchar()) r=r*10+c-‘0‘; return k*r; }
inline const int max(const int &a, const int &b) { return a>b?a:b; }
inline const int min(const int &a, const int &b) { return a<b?a:b; }

const double eps=1e-6;
int dcmp(double x) { return abs(x)<eps?0:(x<0?-1:1); }
struct Point { double x, y; Point(double _x=0, double _y=0) : x(_x), y(_y) {} };

typedef Point Vector;
Vector operator- (Point &a, Point &b) { return Vector(a.x-b.x, a.y-b.y); }
bool operator== (Point &a, Point &b) { return dcmp(a.x-b.x)==0 && dcmp(a.y-b.y)==0; }
double Cross(Vector a, Vector b) { return a.x*b.y-b.x*a.y; }
double Dot(Vector a, Vector b) { return a.x*b.x+a.y*b.y; }
int SSjiao(Point p1, Point p2, Point q1, Point q2) {
	return (dcmp(Cross(p1-q1, q2-q1))^dcmp(Cross(p2-q1, q2-q1)))==-2 &&
		   (dcmp(Cross(q1-p1, p2-p1))^dcmp(Cross(q2-p1, p2-p1)))==-2;
}
int onSegment(Point a, Point b, Point c) {
	if(a==b || a==c) return -1;
	if(dcmp(Cross(a-b, c-b))==0 && dcmp(Dot(b-a, c-a))==-1) return 1;
	return 0;
}

const int N=100015;
int nxt[N], ans[N];
Point a[N][2];

bool check(int now, int goal) {
	if(onSegment(a[now][0], a[goal][0], a[goal][1]) ||
	   onSegment(a[now][1], a[goal][0], a[goal][1]) ||
	   onSegment(a[goal][0], a[now][0], a[now][1])  ||
	   onSegment(a[goal][1], a[now][0], a[now][1])  ) return 1;
	if(SSjiao(a[now][0], a[now][1], a[goal][0], a[goal][1])) return 1;
	return 0;
}
void del(int f, int now) { nxt[f]=nxt[now]; }
void work(int goal) {
	int now=nxt[0], fa=0;
	while(now!=goal) {
		if(check(now, goal)) del(fa, now);
		else fa=now;
		now=nxt[now];
	}
}

int main() {
	int n;
	while(read(n), n) {
		nxt[0]=1;
		for1(i, 1, n) nxt[i]=i+1;
		for1(i, 1, n) {
			rep(k, 2) scanf("%lf%lf", &a[i][k].x, &a[i][k].y);
			work(i);
		}
		printf("Top sticks: ");
		int now=0, cnt=0;
		while(nxt[now]!=n+1) {
			ans[++cnt]=nxt[now];
			now=nxt[now];
		}
		for1(i, 1, cnt-1) printf("%d, ", ans[i]); printf("%d.\n", ans[cnt]);
	}
	return 0;
}

5
1 1 4 2
2 3 3 1
1 -2.0 8 4
1 4 8 2
3 3 6 -2.0
3
0 0 1 1
1 0 2 1
2 0 3 1
0

Top sticks: 2, 4, 5.
Top sticks: 1, 2, 3.