简单几何(线段相交)+模拟 POJ 3449 Geometric Shapes
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。
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247 248 249 250 251 252 253 254 255 256 257 258 259 260 261 262 263 | /************************************************ * 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; } |
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