【POJ】1279 Art Gallery

http://poj.org/problem?id=1279

题意：给一个n个点的多边形，n<=1500，求在多边形内能看到所有多边形上的点的面积。

#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 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 rdm(x, i) for(int i=ihead[x]; i; i=e[i].next)
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; }

const double eps=1e-6;
const int N=1515;
int dcmp(double x) { return abs(x)<eps?0:(x<0?-1:1); }
struct iP { double x, y; iP(double _x=0, double _y=0) : x(_x), y(_y) {} };
typedef iP iV;
iV operator - (iP a, iP b) { return iV(a.x-b.x, a.y-b.y); }
iP operator + (iP a, iV b) { return iP(a.x+b.x, a.y+b.y); }
iV operator * (iP a, double d) { return iV(a.x*d, a.y*d); }
double cross(iV a, iV b) { return a.x*b.y-a.y*b.x; }
double angle(iV &a) { return atan2(a.y, a.x); }

struct iL {
	iV v; iP p;
	double ang;
	void set(iP a, iP b) { p=a; v=b-a; ang=angle(v); }
	bool operator<(const iL &b) const { return ang<b.ang; }
};
iP LLi(iL &a, iL &b) {
	static iV u;
	static double t;
	u=a.p-b.p;
	t=cross(b.v, u)/cross(a.v, b.v);
	return a.p+a.v*t;
}
bool onL(iP &a, iL &l) { return dcmp(cross(l.v, a-l.p))>0; }
bool half(iL *line, int n, iP *s, int &cnt) {
	static iL a[N], q[N];
	static iP b[N];
	static int front, tail;
	memcpy(a, line, sizeof(iL)*(n+1));
	sort(a+1, a+1+n);
	q[front=tail=0]=a[1];
	for1(i, 2, n) {
		while(front!=tail && !onL(b[tail-1], a[i])) --tail;
		while(front!=tail && !onL(b[front], a[i])) ++front;
		q[++tail]=a[i];
		if(dcmp(cross(q[tail-1].v, q[tail].v))==0) {
			--tail;
			if(onL(a[i].p, q[tail])) q[tail]=a[i];
		}
		if(front!=tail) b[tail-1]=LLi(q[tail], q[tail-1]);
	}
	while(front!=tail && !onL(b[tail-1], q[front])) --tail;
	if(tail-front<=1) return 0;
	cnt=0;
	b[tail]=LLi(q[tail], q[front]);
	for1(i, front, tail) s[++cnt]=b[i];
	return 1;
}

iL line[N];
iP a[N], b[N];
int ln, n, num, flag;
void add(iP a, iP b) { ++ln; line[ln].set(a, b); }

void clr() { num=ln=0; }
void readin() { read(n); for1(i, 1, n) scanf("%lf%lf", &a[i].x, &a[i].y); }
void build() {
	a[n+1]=a[1];
	for1(i, 1, n) add(a[i+1], a[i]);
}
void work() { flag=half(line, ln, b, num); }
void getans() {
	if(!flag) { puts("0.00"); return; }
	double ans=0;
	b[num+1]=b[1];
	for1(i, 1, num) ans+=b[i].x*b[i+1].y-b[i].y*b[i+1].x;
	printf("%.2f
", ans/2);
}

int main() {
	int ta=getint();
	while(ta--) {
		clr();
		readin();
		build();
		work();
		getans();
	}
	return 0;
}

　　

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裸的半平面交.................