POJ 2653Pick-up sticks 判断线段相交
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一定要注意位运算的优先级!!!我被这个卡了好久
判断线段相交模板题。
叉积,点积,规范相交,非规范相交的简单模板
用了“链表”优化之后还是$O(n^2)$的暴力,可是为什么能过$10^5$的数据?
#include<cmath> #include<cstdio> #include<cstring> #include<algorithm> #define N 100005 using namespace std; struct Point { double x, y; Point(double _x = 0, double _y = 0) : x(_x), y(_y) {} }; inline int dcmp(double x) { return fabs(x) < 1e-6 ? 0 : (x < 0 ? -1 : 1); } Point operator - (Point a, Point b) { return Point(a.x - b.x, a.y - b.y); } bool operator == (Point a, Point b) { return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0; } double Cross(Point a, Point b) { return a.x * b.y - a.y * b.x; } double Dot(Point a, Point b) { return a.x * b.x + a.y * b.y; } bool jiao(Point d1, Point d2, Point d3, Point d4) { return (dcmp(Cross(d4 - d3, d1 - d3)) ^ dcmp(Cross(d4 - d3, d2 - d3))) == -2 && (dcmp(Cross(d2 - d1, d3 - d1)) ^ dcmp(Cross(d2 - d1, d4 - d1))) == -2; } int bjiao(Point d1, Point d2, Point d3) { if (d1 == d2 || d1 == d3) return 1; if (dcmp(Cross(d2 - d1, d3 - d1)) == 0 && dcmp(Dot(d2 - d1, d3 - d1)) == -1) return 1; return 0; } Point d[N][2]; int n, next[N], ans[N], cnt; inline bool pd(int now, int up) { if (bjiao(d[now][0], d[up][0], d[up][1]) || bjiao(d[now][1], d[up][0], d[up][1]) || bjiao(d[up][0], d[now][0], d[now][1]) || bjiao(d[up][1], d[now][0], d[now][1])) return 1; return jiao(d[now][0], d[now][1], d[up][0], d[up][1]); } inline void mktb(int up) { for(int now = next[0], pre = 0; now != up; now = next[now]) { if (pd(now, up)) next[pre] = next[now]; else pre = now; } } int main() { scanf("%d", &n); while (n) { for(int i = 0; i <= n; ++i) next[i] = i + 1; for(int i = 1; i <= n; ++i) { scanf("%lf%lf%lf%lf", &d[i][0].x, &d[i][0].y, &d[i][1].x, &d[i][1].y); mktb(i); } cnt = 0; for(int now = next[0]; now != n + 1; now = next[now]) ans[++cnt] = now; printf("Top sticks:"); for(int i = 1; i < cnt; ++i) printf("% d,", ans[i]); printf(" %d.\n", ans[cnt]); scanf("%d", &n); } return 0; }
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