标签:return ros next case 二进制 sort pre cross namespace
思路:
二进制枚举一下要删哪些点
求个凸包,算一下贡献
//By SiriusRen #include <cmath> #include <cstdio> #include <algorithm> using namespace std; #define eps 1e-9 int n,cases,top,tot,k,rem,ansrem,ans; double length,tempdis,wei,answei,remwei; struct Tree{double x,y,v,l;}tr[25],point[25],tubao[25]; bool operator<(Tree a,Tree b){if(abs(a.x-b.x)<eps)return a.y<b.y;return a.x<b.x;} double operator*(Tree a,Tree b){return a.x*b.y-a.y*b.x;} Tree operator-(Tree a,Tree b){Tree c;c.x=a.x-b.x;c.y=a.y-b.y;return c;} double dis(Tree a){return sqrt(a.x*a.x+a.y*a.y);} double cross(Tree a,Tree b,Tree c){return (a-c)*(b-c);} int main(){ while(scanf("%d",&n)&&n){ answei=1000000; for(int i=0;i<n;i++)scanf("%lf%lf%lf%lf",&tr[i].x,&tr[i].y,&tr[i].v,&tr[i].l); for(int i=0;i<(1<<n);i++){ length=tot=top=rem=tempdis=wei=0; for(int j=0;j<n;j++){ if(i&(1<<j))point[++tot]=tr[j],rem++; else length+=tr[j].l,wei+=tr[j].v; } sort(point+1,point+1+tot); for(int j=1;j<=tot;j++){ while(top>1&&cross(tubao[top],point[j],tubao[top-1])<-eps)top--; tubao[++top]=point[j]; }k=top; for(int j=tot-1;j>=1;j--){ while(top>k&&cross(tubao[top],point[j],tubao[top-1])<-eps)top--; tubao[++top]=point[j]; } for(int j=1;j<top;j++)tempdis+=dis(tubao[j]-tubao[j+1]); if(tempdis<length){ if(wei<answei)answei=wei,ansrem=rem,ans=i,remwei=length-tempdis; else if(wei==answei&&rem<ansrem)ansrem=rem,ans=i,remwei=length-tempdis; } } printf("Forest %d\nCut these trees:",++cases); for(int j=0;j<n;j++)if(!(ans&(1<<j)))printf(" %d",j+1); printf("\nExtra wood: %.2lf\n\n",remwei); } }
标签:return ros next case 二进制 sort pre cross namespace
原文地址:https://www.cnblogs.com/SiriusRen/p/9383102.html