题意:求多边形的核的面积。
敲一下半平面交模板........ 然后我wa了一早上就因为写了%lf 不知道poj什么破机制还不能用lf的,真的想跳楼
#include<iostream> #include<cstdio> #include<cmath> #include<algorithm> #define MN 50000 #define eps 1e-10 using namespace std; inline int read() { int x = 0 , f = 1; char ch = getchar(); while(ch < '0' || ch > '9'){ if(ch == '-') f = -1; ch = getchar();} while(ch >= '0' && ch <= '9'){x = x * 10 + ch - '0';ch = getchar();} return x * f; } struct P{ double x,y; P(double _x=0,double _y=0):x(_x),y(_y){} P operator + (P b) {return P(x+b.x,y+b.y);} P operator - (P b) {return P(x-b.x,y-b.y);} P operator * (double b) {return P(x*b,y*b);} friend double cross(P a,P b) {return a.x*b.y-b.x*a.y;} }p[MN+5],pt[MN+5]; struct L{ P p,v;double slop; L(){} L(P x,P y):p(x),v(y){slop=atan2(y.y,y.x);} P operator * (L b){P a=p-b.p;double t=cross(b.v,a)/cross(v,b.v);return p+v*t;} bool left(P b){ return cross(v,b-p)>eps;} bool operator < (const L &y) const {return slop<y.slop;} }s[MN+5],q[MN+5]; int n,top,tail; void solve() { q[top=tail=1]=s[1]; for(int i=2;i<=n;i++) { while(top>tail&&!s[i].left(p[top])) --top; while(top>tail&&!s[i].left(p[tail+1])) ++tail; if(fabs(s[i].slop-q[top].slop)<eps) q[top]=s[i].left(q[top].p)?q[top]:s[i]; else q[++top]=s[i]; p[top]=q[top]*q[top-1]; } while(tail<top&&(q[tail].left(p[top])==0)) --top; } int main() { for(int T=read();T;T--) { n=read(); for(int i=1;i<=n;i++) scanf("%lf%lf",&pt[i].x,&pt[i].y); for(int i=2;i<=n;i++) s[i]=L(pt[i-1],pt[i]-pt[i-1]); s[1]=L(pt[n],pt[1]-pt[n]); sort(s+1,s+n+1); solve(); if(top-tail+1<3) { for(int i=1;i<n;i++) s[i]=L(pt[i+1],pt[i]-pt[i+1]); s[n]=L(pt[1],pt[n]-pt[1]); sort(s+1,s+n+1);solve(); } if(top-tail+1<3) puts("0.00"); else { p[tail]=q[tail]*q[top];double area=0; for(int i=tail;i<top;i++) area+=cross(p[i],p[i+1]); area+=cross(p[top],p[tail]); printf("%.2f ",fabs(area)/2.00+eps); } } return 0; }