http://poj.org/problem?id=3525
题目大意：给一个逆时针序列的多边形点集，求其中可以画的最大半径的圆的半径。
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二分枚举半径长度，然后将所有的边往内缩半径为r，求是否有内核即可。
#include<cstdio> #include<queue> #include<cctype> #include<cstring> #include<stack> #include<cmath> #include<algorithm> using namespace std; typedef double dl; const dl eps=1e-10; const int N=301; struct Point{ dl x; dl y; }p[N],point[N],q[N],z; //point，初始点 //q，暂时存可行点 //p，记录可行点 int n,curcnt,cnt; //curcnt，暂时存可行点个数 //cnt，记录可行点个数 inline Point getmag(Point a,Point b){ Point s; s.x=b.x-a.x;s.y=b.y-a.y; return s; } inline dl multiX(Point a,Point b){ return a.x*b.y-b.x*a.y; } inline void getline(Point x,Point y,dl &a,dl &b,dl &c){ a=y.y-x.y; b=x.x-y.x; c=y.x*x.y-x.x*y.y; return; } inline Point intersect(Point x,Point y,dl a,dl b,dl c){ Point s; dl u=fabs(a*x.x+b*x.y+c); dl v=fabs(a*y.x+b*y.y+c); s.x=(x.x*v+y.x*u)/(u+v); s.y=(x.y*v+y.y*u)/(u+v); return s; } inline void cut(dl a,dl b,dl c){ curcnt=0; for(int i=1;i<=cnt;i++){ if(a*p[i].x+b*p[i].y+c>-eps)q[++curcnt]=p[i]; else{ if(a*p[i-1].x+b*p[i-1].y+c>eps){ q[++curcnt]=intersect(p[i],p[i-1],a,b,c); } if(a*p[i+1].x+b*p[i+1].y+c>eps){ q[++curcnt]=intersect(p[i],p[i+1],a,b,c); } } } for(int i=1;i<=curcnt;i++)p[i]=q[i]; p[curcnt+1]=p[1];p[0]=p[curcnt]; cnt=curcnt; return; } inline void init(){ for(int i=1;i<=n;i++)p[i]=point[i]; z.x=z.y=0; p[n+1]=p[1]; p[0]=p[n]; cnt=n; return; } inline void regular(){//调换方向 for(int i=1;i<(n+1)/2;i++)swap(point[i],point[n-i]); return; } inline bool solve(dl r){ init(); for(int i=1;i<=n;i++){ Point ta,tb,tt; tt.x=point[i+1].y-point[i].y; tt.y=point[i].x-point[i+1].x; dl k=r/sqrt(tt.x*tt.x+tt.y*tt.y); tt.x*=k;tt.y*=k; ta.x=point[i].x+tt.x; ta.y=point[i].y+tt.y; tb.x=point[i+1].x+tt.x; tb.y=point[i+1].y+tt.y; dl a,b,c; getline(ta,tb,a,b,c); cut(a,b,c); } return cnt; } int main(){ while(scanf("%d",&n)!=EOF&&n){ for(int i=1;i<=n;i++){ scanf("%lf%lf",&point[i].x,&point[i].y); } regular(); point[n+1]=point[1]; dl l=0,r=10000000; while(fabs(l-r)>eps){ dl mid=(l+r)/2.0; if(solve(mid))l=mid; else r=mid; } printf("%.6f\n",l); } return 0; }