【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\n", ans/2); } int main() { int ta=getint(); while(ta--) { clr(); readin(); build(); work(); getans(); } return 0; }
裸的半平面交.................
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