UVa 361 - Cops and Robbers

题目：官兵与盗贼0 0？在二维平面上给出c个官兵、r个盗贼和o个市民，市民如果在3个官兵包围的三角形中状态是safe，如果在不safe的情况下被3个盗贼包围的三角形中状态是robbed，其他情况状态是neither，现在要求输出所有市民状态。

分析：计算几何、凸包、点与多边形关系。方案1：暴力，显然会超时。方案2：凸包，我们可以得到结论，如果一个点在另一个点集中的三个点的构成三角形中的充要条件就是它在这个点集的凸包中（只要画一画就明白了）。所以只要分别计算官兵和盗贼给成的凸包，然后判断每个市民与两个凸包的关系即可。点与多边形位置关系判断：将点与无限远处的点连线判断射线和多边形交点个数即可，多取几个点可防止出错，所有情况下交点个数均为奇数就是在里面。如过在边上需要特判。

注意：三个相同的点可以构成三角形，即三角形可以是线段或点，但是点的个数必须>=3。

#include <algorithm>#include <iostream>#include <cstdlib>#include <cstdio>#include <cmath>using namespace std;//点结构 typedef struct pnode{double x,y,d;pnode( double a, double b ){x = a;y = b;}pnode(){}}point;point P0,C[205],R[205],O[205];//线结构 typedef struct lnode{double x,y,dx,dy;lnode( point a, point b ){x = a.x;y = a.y;dx = b.x-a.x;dy = b.y-a.y;}lnode(){}}line; //叉乘ab*ac double crossproduct( point a, point b, point c ){return (b.x-a.x)*(c.y-a.y)-(c.x-a.x)*(b.y-a.y);}//点到点距离 double dist( point a, point b ){return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);}//坐标排序 bool cmp1( point a, point b ){return (a.x==b.x)?(a.y<b.y):(a.x<b.x);}//级角排序 bool cmp2( point a, point b ){double cp = crossproduct( P0, a, b );if ( cp == 0 ) return a.d < b.d;else return cp > 0;}//计算凸包 int Graham( point *P, int n ){if ( n < 3 ) return n;sort( P+0, P+n, cmp1 );for ( int i = 1 ; i < n ; ++ i )P[i].d = dist( P[0], P[i] );P0 = P[0];sort( P+1, P+n, cmp2 );int top = n;if ( n > 2 ) {top = 1;for ( int i = 2 ; i < n ; ++ i ) {while ( top > 0 && crossproduct( P[top-1], P[top], P[i] ) <= 0 ) -- top;P[++ top] = P[i];}P[++ top] = P[0];}return top;}//点在直线上 bool on( point p, line l ){if ( l.dx*(p.y-l.y)-l.dy*(p.x-l.x) == 0 )if ( (p.x-l.x)*(p.x-l.x-l.dx) <= 0 )if ( (p.y-l.y)*(p.y-l.y-l.dy) <= 0 )return true;return false;}//线段相交 bool cross( line a, line b ){double t1 = 0.0+a.dx*(b.y-a.y)-a.dy*(b.x-a.x);double t2 = 0.0+a.dx*(b.y+b.dy-a.y)-a.dy*(b.x+b.dx-a.x);double t3 = 0.0+b.dx*(a.y-b.y)-b.dy*(a.x-b.x);double t4 = 0.0+b.dx*(a.y+a.dy-b.y)-b.dy*(a.x+a.dx-b.x);return (t1*t2 <= 0)&&(t3*t4 <= 0);}//点与多边形关系判断bool in( point p, point* P, int n ){double d[4][2] = {-505,-505,-505,505,505,-505,505,505};for ( int t = 0 ; t < 4 ; ++ t ) {line s1 = line( p, point( d[t][0], d[t][1] ) );int  count = 0;for ( int i = 0 ; i < n ; ++ i ) {line s2 = line( P[i], P[i+1] );if ( on( p, s2 ) ) return true;if ( cross( s1, s2 ) ) count ++;if ( on( P[i], s1 ) ) count --;}if ( count%2 == 0 ) return false; }return true;}int main(){int c,r,o,t = 1;while ( scanf("%d%d%d",&c,&r,&o) && c+r+o ) {for ( int i = 0 ; i < c ; ++ i ) scanf("%lf%lf",&C[i].x,&C[i].y);for ( int i = 0 ; i < r ; ++ i )scanf("%lf%lf",&R[i].x,&R[i].y);for ( int i = 0 ; i < o ; ++ i )scanf("%lf%lf",&O[i].x,&O[i].y);int c_count = Graham( C, c );int r_count = Graham( R, r );printf("Data set %d:\n",t ++);for ( int i = 0 ; i < o ; ++ i ) {printf("     Citizen at (%.0lf,%.0lf) is ",O[i].x,O[i].y);if ( c > 2 && in( O[i], C, c_count ) ) printf("safe.\n");else if ( r > 2 && in( O[i], R, r_count ) ) printf("robbed.\n");else printf("neither.\n");}printf("\n");}return 0;}