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题意:给了若干个图形,问每个图形与哪些图形相交
分析:题目说白了就是处理出每个图形的线段,然后判断是否相交。但是读入输出巨恶心,就是个模拟题加上线段相交的判断,我第一次WA不知道输出要按字母序输出,第二次WA是因为忘记多边形的最后一条线段,还好找到了,没有坚持的话就不会AC了。
/************************************************ * Author :Running_Time * Created Time :2015/10/31 星期六 13:38:11 * File Name :POJ_3449.cpp ************************************************/ #include <cstdio> #include <algorithm> #include <iostream> #include <sstream> #include <cstring> #include <cmath> #include <string> #include <vector> #include <queue> #include <deque> #include <stack> #include <list> #include <map> #include <set> #include <bitset> #include <cstdlib> #include <ctime> using namespace std; #define lson l, mid, rt << 1 #define rson mid + 1, r, rt << 1 | 1 typedef long long ll; const int N = 1e5 + 10; const int INF = 0x3f3f3f3f; const int MOD = 1e9 + 7; const double EPS = 1e-10; const double PI = acos (-1.0); int dcmp(double x) { //三态函数,减少精度问题 if (fabs (x) < EPS) return 0; else return x < 0 ? -1 : 1; } struct Point { //点的定义 double x, y; Point () {} Point (double x, double y) : x (x), y (y) {} Point operator + (const Point &r) const { //向量加法 return Point (x + r.x, y + r.y); } Point operator - (const Point &r) const { //向量减法 return Point (x - r.x, y - r.y); } Point operator * (double p) const { //向量乘以标量 return Point (x * p, y * p); } Point operator / (double p) const { //向量除以标量 return Point (x / p, y / p); } bool operator < (const Point &r) const { //点的坐标排序 return x < r.x || (x == r.x && y < r.y); } bool operator == (const Point &r) const { //判断同一个点 return dcmp (x - r.x) == 0 && dcmp (y - r.y) == 0; } }; typedef Point Vector; //向量的定义 Point read_point(void) { //点的读入 double x, y; scanf ("%lf%lf", &x, &y); return Point (x, y); } double dot(Vector A, Vector B) { //向量点积 return A.x * B.x + A.y * B.y; } double cross(Vector A, Vector B) { //向量叉积 return A.x * B.y - A.y * B.x; } double polar_angle(Vector A) { //向量极角 return atan2 (A.y, A.x); } double length(Vector A) { //向量长度,点积 return sqrt (dot (A, A)); } double angle(Vector A, Vector B) { //向量转角,逆时针,点积 return acos (dot (A, B) / length (A) / length (B)); } Vector rotate(Vector A, double rad) { //向量旋转,逆时针 return Vector (A.x * cos (rad) - A.y * sin (rad), A.x * sin (rad) + A.y * cos (rad)); } Vector nomal(Vector A) { //向量的单位法向量 double len = length (A); return Vector (-A.y / len, A.x / len); } Point line_line_inter(Point p, Vector V, Point q, Vector W) { //两直线交点,参数方程 Vector U = p - q; double t = cross (W, U) / cross (V, W); return p + V * t; } double point_to_line(Point p, Point a, Point b) { //点到直线的距离,两点式 Vector V1 = b - a, V2 = p - a; return fabs (cross (V1, V2)) / length (V1); } double point_to_seg(Point p, Point a, Point b) { //点到线段的距离,两点式 if (a == b) return length (p - a); Vector V1 = b - a, V2 = p - a, V3 = p - b; if (dcmp (dot (V1, V2)) < 0) return length (V2); else if (dcmp (dot (V1, V3)) > 0) return length (V3); else return fabs (cross (V1, V2)) / length (V1); } Point point_line_proj(Point p, Point a, Point b) { //点在直线上的投影,两点式 Vector V = b - a; return a + V * (dot (V, p - a) / dot (V, V)); } bool can_seg_seg_inter(Point a1, Point a2, Point b1, Point b2) { //判断线段相交,两点式 double c1 = cross (a2 - a1, b1 - a1), c2 = cross (a2 - a1, b2 - a1), c3 = cross (b2 - b1, a1 - b1), c4 = cross (b2 - b1, a2 - b1); return dcmp (c1) * dcmp (c2) <= 0 && dcmp (c3) * dcmp (c4) <= 0; } bool can_line_seg_inter(Point a1, Point a2, Point b1, Point b2) { //判断直线与线段相交,两点式 double c1 = cross (a2 - a1, b1 - a1), c2 = cross (a2 - a1, b2 - a1); return dcmp (c1 * c2) <= 0; } bool on_seg(Point p, Point a1, Point a2) { //判断点在线段上,两点式 return dcmp (cross (a1 - p, a2 - p)) == 0 && dcmp (dot (a1 - p, a2 - p)) < 0; } double area_triangle(Point a, Point b, Point c) { //三角形面积,叉积 return fabs (cross (b - a, c - a)) / 2.0; } double area_poly(Point *p, int n) { //多边形面积,叉积 double ret = 0; for (int i=1; i<n-1; ++i) { ret += fabs (cross (p[i] - p[0], p[i+1] - p[0])); } return ret / 2; } struct Seg { Point a, b; Seg () {} Seg (Point a, Point b) : a (a), b (b) {} }; struct Pic { char ch; int cnt; vector<Seg> seg; vector<int> ans; bool operator < (const Pic &r) const { return ch < r.ch; } }p[33], pp[33]; bool cmp(int i, int j) { return p[i].ch < p[j].ch; } void run(int tot) { for (int i=0; i<tot; ++i) { p[i].cnt = 0; for (int j=0; j<tot; ++j) { if (i == j) continue; int sz1 = p[i].seg.size (), sz2 = p[j].seg.size (); bool flag = false; for (int ii=0; ii<sz1; ++ii) { for (int jj=0; jj<sz2; ++jj) { if (can_seg_seg_inter (p[i].seg[ii].a, p[i].seg[ii].b, p[j].seg[jj].a, p[j].seg[jj].b)) { flag = true; p[i].cnt++; p[i].ans.push_back (j); break; } } if (flag) break; } } sort (p[i].ans.begin (), p[i].ans.end (), cmp); } } int main(void) { //freopen ("POJ_3449.in", "r", stdin); //freopen ("POJ_3449.out", "w", stdout); char str[22]; int x[22], y[22]; int tot = 0; int x1, y1, x2, y2, x3, y3, x4, y4; Point p1, p2, p3, p4; while (scanf ("%s", &str) == 1) { if (strcmp (str, ".") == 0) break; else if (strcmp (str, "-") == 0) { run (tot); for (int i=0; i<tot; ++i) pp[i] = p[i]; sort (p, p+tot); for (int i=0; i<tot; ++i) { if (p[i].cnt == 0) { printf ("%c has no intersections\n", p[i].ch); } else { if (p[i].cnt == 1) { printf ("%c intersects with %c\n", p[i].ch, pp[p[i].ans[0]].ch); } else if (p[i].cnt == 2) { printf ("%c intersects with %c and %c\n", p[i].ch, pp[p[i].ans[0]].ch, pp[p[i].ans[1]].ch); } else { printf ("%c intersects with %c", p[i].ch, pp[p[i].ans[0]].ch); int sz = p[i].ans.size (); for (int j=1; j<sz-1; ++j) { printf (", %c", pp[p[i].ans[j]].ch); } printf (", and %c\n", pp[p[i].ans[sz-1]].ch); } } p[i].ans.clear (); p[i].seg.clear (); } puts (""); tot = 0; continue; } p[tot].ch = str[0]; scanf ("%s", &str); if (str[0] == ‘s‘) { scanf (" (%d,%d) (%d,%d)", &x1, &y1, &x3, &y3); //cal x2 x4 p1 = Point (x1, y1), p3 = Point (x3, y3), p2, p4; Vector V = p3 - p1; double len = length (V); p2 = p1 + rotate (V, PI / 4) / sqrt (2.0); p4 = p1 + rotate (V, -PI / 4) / sqrt (2.0); p[tot].seg.push_back (Seg (p1, p2)); p[tot].seg.push_back (Seg (p2, p3)); p[tot].seg.push_back (Seg (p3, p4)); p[tot].seg.push_back (Seg (p4, p1)); } else if (str[0] == ‘l‘) { scanf (" (%d,%d) (%d,%d)", &x1, &y1, &x2, &y2); p[tot].seg.push_back (Seg (Point (x1, y1), Point (x2, y2))); } else if (str[0] == ‘t‘) { scanf (" (%d,%d) (%d,%d) (%d,%d)", &x1, &y1, &x2, &y2, &x3, &y3); p[tot].seg.push_back (Seg (Point (x1, y1), Point (x2, y2))); p[tot].seg.push_back (Seg (Point (x2, y2), Point (x3, y3))); p[tot].seg.push_back (Seg (Point (x3, y3), Point (x1, y1))); } else if (str[0] == ‘p‘) { int n; scanf ("%d", &n); for (int i=0; i<n; ++i) { scanf (" (%d,%d)", &x[i], &y[i]); } for (int i=0; i<n-1; ++i) { p[tot].seg.push_back (Seg (Point (x[i], y[i]), Point (x[i+1], y[i+1]))); p[tot].seg.push_back (Seg (Point (x[n-1], y[n-1]), Point (x[0], y[0]))); //忘记加了 } } else if (str[0] == ‘r‘) { scanf (" (%d,%d) (%d,%d) (%d,%d)", &x1, &y1, &x2, &y2, &x3, &y3); //cal x4 p1 = Point (x1, y1), p3 = Point (x3, y3), p2 = Point (x2, y2), p4; p4 = p1 + (p3 - p2); p[tot].seg.push_back (Seg (p1, p2)); p[tot].seg.push_back (Seg (p2, p3)); p[tot].seg.push_back (Seg (p3, p4)); p[tot].seg.push_back (Seg (p4, p1)); } tot++; } //cout << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n"; return 0; }
简单几何(线段相交)+模拟 POJ 3449 Geometric Shapes
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原文地址:http://www.cnblogs.com/Running-Time/p/4925663.html