铁轨 (Magnetic Train Tracks,Dhaka 2007,LA 4064)
1 #include <iostream> 2 #include <string.h> 3 #include <string> 4 #include <fstream> 5 #include <algorithm> 6 #include <stdio.h> 7 #include <vector> 8 #include <queue> 9 #include <set> 10 #include <cmath> 11 using namespace std; 12 const double eps = 1e-8; 13 const double pi=acos(-1.0); 14 const int INF=0x7fffffff; 15 unsigned long long uINF = ~0LL; 16 #define MAXN 10000007 17 typedef long long LL; 18 struct node 19 { 20 double x,y; 21 }p[1500]; 22 23 node operator-(node a,node b) 24 { 25 a.x-=b.x; 26 a.y-=b.y; 27 return a; 28 } 29 int check(double x) 30 { 31 if(fabs(x)<eps)return 0; 32 return x<0? -1:1; 33 } 34 double ang[5500]; 35 int main() 36 { 37 int n;int t=1; 38 while(scanf("%d",&n),n) 39 { 40 for(int i=0;i<n;i++) 41 scanf("%lf%lf",&p[i].x,&p[i].y); 42 43 44 printf("Scenario %d:\n",t++); 45 if(n<3) 46 { 47 printf("There are 0 sites for making valid tracks\n"); 48 continue; 49 } 50 int ans=0; 51 for(int i=0;i<n;i++) 52 { 53 int len=0;node temp; 54 for(int j=0;j<n;j++) 55 { 56 if(i==j)continue; 57 temp=p[j]-p[i]; 58 ang[len++]=atan2(temp.y,temp.x); 59 } 60 sort(ang,ang+len); 61 for(int j=0;j<len;j++) 62 ang[j+len]=ang[j]+2*pi; 63 int t=0; 64 for(int j=0;j<len;j++) 65 { 66 while(check(ang[t]-ang[j]-pi/2)<0)t++; 67 ans+=t-j-1; 68 } 69 } 70 printf("There are %d sites for making valid tracks\n",ans-(n*(n-1)*(n-2))/3); 71 } 72 73 return 0; 74 }
统计 2边 角度小于pi/2 的组合 减去重复的 及所求三角形个数
