首页 > 代码库 > 【POJ】1279

【POJ】1279

http://poj.org/submit?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"); }	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\n", ans/2);}int main() {	int ta=getint();	while(ta--) {		clr();		readin();		build();		work();		getans();	}	return 0;}

  


 

裸的半平面交.................

【POJ】1279