#include#include#include#include#include#include#include#include#include#define CL(arr, val) memset(arr, val, sizeof(arr))#define REP(i, n) for((i) = 0; (i) < (n); ++(i))#define FOR(i, l, h) for((i) = (l); (i) <= (h); ++(i))#define FORD(i, h, l) for((i) = (h); (i) >= (l); --(i))#define L(x) (x) << 1#define R(x) (x) << 1 | 1#define MID(l, r) (l + r) >> 1#define Min(x, y) x < y ? x : y#define Max(x, y) x < y ? y : x#define E(x) (1 << (x))const double eps = 1e-8;typedef long long LL;using namespace std;const int maxn = 110;struct Point { double x; double y; Point(double a = 0, double b = 0): x(a), y(b) {} void input() { scanf("%lf%lf", &x, &y); }};Point point[maxn], p[maxn], q[maxn]; //读入的多边形的顶点（顺时针）、p为存放最终切割得到的多边形顶点的数组、暂存核的顶点 int cCnt, n; //此时cCnt为最终切割得到的多边形的顶点数、暂存顶点个数inline int dbcmp(double x) { //精度问题 if(x > eps) return 1; else if(x < -eps) return -1; return 0;}inline void getline(Point x, Point y, double& a, double& b, double& c) { //点X，Y确定一条直线 a = y.y - x.y; b = x.x - y.x; c = y.x*x.y - x.x*y.y;}inline Point intersect(Point x, Point y, double a, double b, double c) { ////求x、y形成的直线与已知直线a、b、c、的交点 double u = fabs(a*x.x + b*x.y + c); double v = fabs(a*y.x + b*y.y + c); return Point((x.x*v + y.x*u)/(u + v), (x.y*v + y.y*u)/(u + v));}inline void cut(double a, double b, double c) { //如上图所示，切割 int cur = 0, i; for(i = 1; i <= cCnt; ++i) { if(dbcmp(a*p[i].x + b*p[i].y + c) >= 0) q[++cur] = p[i]; // c由于精度问题，可能会偏小，所以有些点本应在右侧而没在 else { if(dbcmp(a*p[i-1].x + b*p[i-1].y + c) > 0) //如果p[i-1]在直线的右侧的话， //则将p[i],p[i-1]形成的直线与已知直线的交点作为核的一个顶点(这样的话，由于精度的问题，核的面积可能会有所减少) q[++cur] = intersect(p[i], p[i-1], a, b, c); if(dbcmp(a*p[i+1].x + b*p[i+1].y + c) > 0) q[++cur] = intersect(p[i], p[i+1], a, b, c); } } for(i = 1; i <= cur; ++i) p[i] = q[i]; p[cur+1] = q[1]; p[0] = p[cur]; cCnt = cur;}void solve() { //注意：默认点是顺时针，如果题目不是顺时针，规整化方向 int i; for(i = 1; i <= n; ++i) { double a, b, c; getline(point[i], point[i+1], a, b, c); cut(a, b, c); } if(cCnt == 0) puts("NO"); else puts("YES");}void init() { int i; FOR(i, 1, n) point[i].input(); point[n+1] = point[1]; //初始化p[], cCnt FOR(i, 1, n) p[i] = point[i]; p[n+1] = p[1]; p[0] = p[n]; cCnt = n;}int main() { //freopen("data.in", "r", stdin); int t; scanf("%d", &t); while(t--) { scanf("%d", &n); init(); solve(); } return 0;}*************************************************************************************HDU1474#include #include #include using namespace std; struct Point { double x, y; } point[105]; //原始节点 Point p[105];//保存新切割出的多边形 Point q[105];//临时保存新切割的多边形 int n, m; double a, b, c;//直线 ax + by + c == 0 void lineFromSegment(Point p1, Point p2) { //线段所在直线,返回直线方程的三个系数 a = p2.y - p1.y; b = p1.x - p2.x; c = p2.x * p1.y - p1.x * p2.y; } Point intersect(Point x, Point y) { double u = fabs(a * x.x + b * x.y + c); double v = fabs(a * y.x + b * y.y + c); Point ans; ans.x = (x.x * v + y.x * u) / (u + v); ans.y = (x.y * v + y.y * u) / (u + v); return ans; }//获取直线ax+by+c==0 和点x和y所连直线的交点 int cut() { int np = 0, i; for( i = 0 ; i < m ; i++) { if( p[i].x *a + p[i].y *b + c >= 0) //题目输入是顺时针方向故大于号 { q[np++] = p[i]; }//某点在该直线割后的多边形内，则保存该点 else { if( p[(i + m - 1) % m].x * a + p[(i + m - 1) % m].y * b + c > 0 )// { q[np++] = intersect(p[i], p[(i + m - 1) % m]); } if( p[(i + 1) % m].x * a + p[(i + 1) % m].y * b + c > 0 ) { q[np++] = intersect(p[i], p[(i + 1) % m]); } }//如果该点在外，看是该点的相邻两边是否存在与割线的交点，有交点保存割边上的交点 } for( i = 0 ; i < np ; i++) p[i] = q[i];//更新切割后的多边形顶点 m = np; return 0; } int find_nucleus() { int i, j; for( i = 0; i < n; i++) p[i] = point[i]; m = n; for( i = 0 ; i < n ; i++) { lineFromSegment(point[i], point[(i + 1) % n]); cut(); } return 0; } int main() { int i, k; k = 1; while(scanf("%d", &n), n) { for( i = 0 ; i < n ; i++) scanf("%lf%lf", &point[i].x, &point[i].y); find_nucleus(); printf("Floor #%d\n", k++); printf("%s\n\n", m == 0 ? "Surveillance is impossible." : "Surveillance is possible."); } return 0; } 
