UVA 361 Cops and Robbers
给你警察点，强盗点，公民点。如果一个公民在警察组成的三角形内，那么他SAFE，不在警察内在强盗内，ROBBED，都不在输出NEITHER。
200个点，如果硬枚举的话，很可能超时了，后来想到，只要求出警察组成的凸包和强盗组成的凸包，判断公民是否在这两个凸包内即可。
数据比较阴险，看dis才知道T T 。。。
如果有三点警察共点，这个公民在这个点上的话也是SAFE，两个共点就不行。。。
RANK 67~
#include <queue> #include <stack> #include <math.h> #include <stdio.h> #include <stdlib.h> #include <iostream> #include <limits.h> #include <string.h> #include <string> #include <algorithm> using namespace std; const int MAX = 220; struct point{ double x,y;}; struct segment{ point a,b;}; point c[MAX],r[MAX]; point cc[MAX],rr[MAX]; int t1,t2; const double eps = 1e-6; bool dy(double x,double y) { return x > y + eps;} // x > y bool xy(double x,double y) { return x < y - eps;} // x < y bool dyd(double x,double y) { return x > y - eps;} // x >= y bool xyd(double x,double y) { return x < y + eps;} // x <= y bool dd(double x,double y) { return fabs( x - y ) < eps;} // x == y double disp2p(point a,point b) { return sqrt( ( a.x - b.x ) * ( a.x - b.x ) + ( a.y - b.y ) * ( a.y - b.y ) ); } double crossProduct(point a,point b,point c)//向量 ac 在 ab 的方向 { return (c.x - a.x)*(b.y - a.y) - (b.x - a.x)*(c.y - a.y); } bool cmp(point a,point b) // 排序 { double len = crossProduct(c[0],a,b); if( dd(len,0.0) ) return xy(disp2p(c[0],a),disp2p(c[0],b)); return xy(len,0.0); } void Graham(point c[],int n,point stk[],int &top) { int tmp = 0; for(int i=1; i<n; i++) if( xy(c[i].x,c[tmp].x) || dd(c[i].x,c[tmp].x) && xy(c[i].y,c[tmp].y) ) tmp = i; swap(c[0],c[tmp]); sort(c+1,c+n,cmp); stk[0] = c[0]; stk[1] = c[1]; top = 1; for(int i=2; i<n; i++) { while( xyd( crossProduct(stk[top],stk[top-1],c[i]), 0.0 ) && top >= 1 ) top--; stk[++top] = c[i]; } } bool onSegment(point a, point b, point c) { double maxx = max(a.x,b.x); double maxy = max(a.y,b.y); double minx = min(a.x,b.x); double miny = min(a.y,b.y); if( dd(crossProduct(a,b,c),0.0) && dyd(c.x,minx) && xyd(c.x,maxx) && dyd(c.y,miny) && xyd(c.y,maxy) ) return true; return false; }bool segIntersect(point p1,point p2, point p3, point p4) { double d1 = crossProduct(p3,p4,p1); double d2 = crossProduct(p3,p4,p2); double d3 = crossProduct(p1,p2,p3); double d4 = crossProduct(p1,p2,p4); if( xy(d1 * d2,0.0) && xy(d3 * d4,0.0) ) return true; if( dd(d1,0.0) && onSegment(p3,p4,p1) ) return true;//如果不判端点相交，则下面这四句话不需要 if( dd(d2,0.0) && onSegment(p3,p4,p2) ) return true; if( dd(d3,0.0) && onSegment(p1,p2,p3) ) return true; if( dd(d4,0.0) && onSegment(p1,p2,p4) ) return true; return false; } bool point_inPolygon(point pot,point p[],int n) { int count = 0; segment l; l.a = pot; l.b.x = 1e10; l.b.y = pot.y; p[n] = p[0]; for(int i=0; i<n; i++) { if( onSegment(p[i],p[i+1],pot) ) return true; if( !dd(p[i].y,p[i+1].y) ) { int tmp = -1; if( onSegment(l.a,l.b,p[i]) ) tmp = i; else if( onSegment(l.a,l.b,p[i+1]) ) tmp = i+1; if( tmp != -1 && dd(p[tmp].y,max(p[i].y,p[i+1].y)) ) count++; else if( tmp == -1 && segIntersect(p[i],p[i+1],l.a,l.b) ) count++; } } if( count % 2 == 1 ) return true; return false; } int main() { int n,m,o; int ind = 1; point pot; while( ~scanf("%d%d%d",&n,&m,&o) ) { t1 = t2 = 0; if( m == n && m == o && o == 0 ) break; for(int i=0; i<n; i++) scanf("%lf%lf",&c[i].x,&c[i].y); for(int i=0; i<m; i++) scanf("%lf%lf",&r[i].x,&r[i].y); if( n ) Graham(c,n,cc,t1); memcpy(c,r,sizeof(r)); if( m ) Graham(c,m,rr,t2); printf("Data set %d:/n",ind++); bool flag1 = false,flag2 = false; if( t1 + 1 <= 2 && n <= 2 ) flag1 = true; if( t2 + 1 <= 2 && m <= 2 ) flag2 = true; while( o-- ) { scanf("%lf%lf",&pot.x,&pot.y); if( !flag1 && point_inPolygon(pot,cc,t1+1) ) { printf(" Citizen at (%.0lf,%.0lf) is safe./n",pot.x,pot.y); continue; } if( !flag2 && point_inPolygon(pot,rr,t2+1) ) { printf(" Citizen at (%.0lf,%.0lf) is robbed./n",pot.x,pot.y); continue; } printf(" Citizen at (%.0lf,%.0lf) is neither./n",pot.x,pot.y); } printf("/n"); } return 0; }