Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions:2105 | Accepted: 883 |
Description
While creating a customer logo, ACM uses graphical utilities to draw a picture that can later be cut into special fluorescent materials. To ensure proper processing, the shapes in the picture cannot intersect. However, some logos contain such intersecting shapes. It is necessary to detect them and decide how to change the picture.
Given a set of geometric shapes, you are to determine all of their intersections. Only outlines are considered, if a shape is completely inside another one, it is not counted as an intersection.
Input
Input contains several pictures. Each picture describes at most 26 shapes, each specified on a separate line. The line begins with an uppercase letter that uniquely identifies the shape inside the corresponding picture. Then there is a kind of the shape and two or more points, everything separated by at least one space. Possible shape kinds are:
• square: Followed by two distinct points giving the opposite corners of the square.
• rectangle: Three points are given, there will always be a right angle between the lines connecting the first point with the second and the second with the third.
• line: Specifies a line segment, two distinct end points are given.
• triangle: Three points are given, they are guaranteed not to be co-linear.
• polygon: Followed by an integer number N (3 ≤ N ≤ 20) and N points specifying vertices of the polygon in either clockwise or anti-clockwise order. The polygon will never intersect itself and its sides will have non-zero length.
All points are always given as two integer coordinates X and Y separated with a comma and enclosed in parentheses. You may assume that |X|, |Y | ≤ 10000.
The picture description is terminated by a line containing a single dash (“-”). After the last picture, there is a line with one dot (“.”).
Output
For each picture, output one line for each of the shapes, sorted alphabetically by its identifier (X). The line must be one of the following:
• “X has no intersections”, if X does not intersect with any other shapes.
• “X intersects with A”, if X intersects with exactly 1 other shape.
• “X intersects with A and B”, if X intersects with exactly 2 other shapes.
• “X intersects with A, B, . . ., and Z”, if X intersects with more than 2 other shapes.
Please note that there is an additional comma for more than two intersections. A, B, etc. are all intersecting shapes, sorted alphabetically.
Print one empty line after each picture, including the last one.
Sample Input
A square (1,2) (3,2) F line (1,3) (4,4) W triangle (3,5) (5,5) (4,3) X triangle (7,2) (7,4) (5,3) S polygon 6 (9,3) (10,3) (10,4) (8,4) (8,1) (10,2) B rectangle (3,3) (7,5) (8,3) - B square (1,1) (2,2) A square (3,3) (4,4) - .
Sample Output
A has no intersections B intersects with S, W, and X F intersects with W S intersects with B W intersects with B and F X intersects with B A has no intersections B has no intersections
Source
题意:
给定一些图形及其坐标(可能是正方形、矩形、三角形、多边形、线段),问每一个图形和其他哪些图形相交
思路:
因为数据量很小,暴力判断每一个图形的每一条边和其他每一个图形的每一条边是否相交
输入输出有点恶心 没别的了
1 //#include <bits/stdc++.h> 2 #include<iostream> 3 #include<cmath> 4 #include<algorithm> 5 #include<stdio.h> 6 #include<cstring> 7 8 using namespace std; 9 typedef long long int LL; 10 11 #define zero(x) (((x) > 0? (x) : -(x)) < eps) 12 //enum shape{"square", "rectangle", "line", "triangle", "polygon"}; 13 const double eps = 1e-8; 14 const double pi = 3.141592654; 15 struct point{ 16 int x,y; 17 point() 18 { 19 x = 0; 20 y = 0; 21 } 22 point(int _x, int _y) 23 { 24 x = _x; 25 y = _y; 26 } 27 }; 28 29 struct node{ 30 int sh; 31 int nvertices; 32 point p[25]; 33 char name; 34 /*bool operator < (const node &b)const 35 { 36 name < b.name; 37 }*/ 38 }; 39 node shapes[30]; 40 41 bool cmp(node a, node b) 42 { 43 return a.name < b.name; 44 } 45 46 double xmult(point p1, point p2, point p0) 47 { 48 return (p1.x - p0.x) * (p2.y - p0.y) - (p2.x - p0.x) * (p1.y - p0.y); 49 } 50 51 //判断三点共线 52 int dots_inline(point p1, point p2, point p3) 53 { 54 return zero(xmult(p1, p2, p3)); 55 } 56 57 //判断点是否在线段上,包括端点 58 int dot_online_in(point p, point l1, point l2) 59 { 60 return zero(xmult(p, l1, l2)) && (l1.x - p.x) * (l2.x - p.x) < eps && (l1.y - p.y) * (l2.y - p.y) < eps; 61 } 62 63 //判断两点在线段同侧,点在线段上返回0 64 int same_side(point p1, point p2, point l1, point l2) 65 { 66 return xmult(l1, p1, l2) * xmult(l1, p2, l2) > eps; 67 } 68 69 //判断两线段相交,包括端点和部分重合 70 int intersect_in(point u1, point u2, point v1, point v2) 71 { 72 if(!dots_inline(u1, u2, v1) || !dots_inline(u1, u2, v2)){ 73 return !same_side(u1, u2, v1, v2) && !same_side(v1, v2, u1, u2); 74 } 75 76 return dot_online_in(u1, v1, v2) || dot_online_in(u2, v1, v2) || dot_online_in(v1, u1, u2) || dot_online_in(v2, u1, u2); 77 78 } 79 80 /*bool inter(point u1, point u2, point v1, point v2) 81 { 82 return 83 max(u1.x, u2.x) >= min(v1.x, v2.x) && 84 max(v1.x, v2.x) >= min(u1.x, u2.x) && 85 max(u1.y, u2.y) >= min(v1.y, v2.y) && 86 max(v1.y, v2.y) >= min(u1.y, u2.y) && 87 sgn((v)) 88 }*/ 89 90 char str[30]; 91 bool inter[30]; 92 int main() 93 { 94 int cnt = 0; 95 while(scanf("%s", str) == 1){ 96 cnt = 0; 97 if(str[0] =='.')break; 98 shapes[0].name = str[0]; 99 scanf("%s", str); 100 if(strcmp(str, "square") == 0){ 101 shapes[0].nvertices = 4; 102 int x0, x2, y0, y2; 103 //cin>>x0>>y0>>x2>>y2; 104 scanf(" (%d,%d)", &x0, &y0); 105 scanf(" (%d,%d)", &x2, &y2); 106 shapes[0].p[0] = point(x0, y0); 107 shapes[0].p[2] = point(x2, y2); 108 shapes[0].p[1] = point((x0 + x2 + y2 - y0) / 2, (y0 + y2 + x0 - x2) / 2); 109 shapes[0].p[3] = point((x0 + x2 - y2 + y0) / 2, (y0 + y2 - x0 + x2) / 2); 110 } 111 else if(strcmp(str, "line") == 0){ 112 shapes[0].nvertices = 2; 113 int x0, y0, x1, y1; 114 scanf(" (%d,%d) (%d,%d)", &x0, &y0, &x1, &y1); 115 shapes[0].p[0] = point(x0, y0); 116 shapes[0].p[1] = point(x1, y1); 117 } 118 else if(strcmp(str, "triangle") == 0){ 119 shapes[0].nvertices = 3; 120 int x0, y0, x1, y1, x2, y2; 121 scanf(" (%d,%d) (%d,%d) (%d,%d)", &x0, &y0, &x1, &y1, &x2, &y2); 122 shapes[0].p[0] = point(x0, y0); 123 shapes[0].p[1] = point(x1, y1); 124 shapes[0].p[2] = point(x2, y2); 125 } 126 else if(strcmp(str, "rectangle") == 0){ 127 shapes[0].nvertices = 4; 128 int x0, y0, x1, y1, x2, y2; 129 scanf(" (%d,%d) (%d,%d) (%d,%d)", &x0, &y0, &x1, &y1, &x2, &y2); 130 shapes[0].p[0] = point(x0, y0); 131 shapes[0].p[1] = point(x1, y1); 132 shapes[0].p[2] = point(x2, y2); 133 shapes[0].p[3] = point(x2 + x0 - x1, y2 + y0 - y1); 134 } 135 else if(strcmp(str, "polygon") == 0){ 136 scanf("%d", &shapes[0].nvertices); 137 for(int i = 0; i < shapes[0].nvertices; i++){ 138 scanf(" (%d,%d)", &shapes[0].p[i].x, &shapes[0].p[i].y); 139 } 140 } 141 cnt++; 142 while(scanf("%s", str) == 1){ 143 if(str[0] == '-')break; 144 shapes[cnt].name = str[0]; 145 scanf("%s", str); 146 if(strcmp(str, "square") == 0){ 147 shapes[cnt].nvertices = 4; 148 int x0, x2, y0, y2; 149 scanf(" (%d,%d) (%d,%d)", &x0, &y0, &x2, &y2); 150 shapes[cnt].p[0] = point(x0, y0); 151 shapes[cnt].p[2] = point(x2, y2); 152 shapes[cnt].p[1] = point((x0 + x2 + y2 - y0) / 2, (y0 + y2 + x0 - x2) / 2); 153 shapes[cnt].p[3] = point((x0 + x2 - y2 + y0) / 2, (y0 + y2 - x0 + x2) / 2); 154 } 155 else if(strcmp(str, "line") == 0){ 156 shapes[cnt].nvertices = 2; 157 int x0, y0, x1, y1; 158 scanf(" (%d,%d) (%d,%d)", &x0, &y0, &x1, &y1); 159 shapes[cnt].p[0] = point(x0, y0); 160 shapes[cnt].p[1] = point(x1, y1); 161 } 162 else if(strcmp(str, "triangle") == 0){ 163 shapes[cnt].nvertices = 3; 164 int x0, y0, x1, y1, x2, y2; 165 scanf(" (%d,%d) (%d,%d) (%d,%d)", &x0, &y0, &x1, &y1, &x2, &y2); 166 shapes[cnt].p[0] = point(x0, y0); 167 shapes[cnt].p[1] = point(x1, y1); 168 shapes[cnt].p[2] = point(x2, y2); 169 } 170 else if(strcmp(str, "rectangle") == 0){ 171 shapes[cnt].nvertices = 4; 172 int x0, y0, x1, y1, x2, y2; 173 scanf(" (%d,%d) (%d,%d) (%d,%d)", &x0, &y0, &x1, &y1, &x2, &y2); 174 shapes[cnt].p[0] = point(x0, y0); 175 shapes[cnt].p[1] = point(x1, y1); 176 shapes[cnt].p[2] = point(x2, y2); 177 shapes[cnt].p[3] = point(x2 + x0 - x1, y2 + y0 - y1); 178 } 179 else if(strcmp(str, "polygon") == 0){ 180 scanf("%d", &shapes[cnt].nvertices); 181 for(int i = 0; i < shapes[cnt].nvertices; i++){ 182 scanf(" (%d,%d)", &shapes[cnt].p[i].x, &shapes[cnt].p[i].y); 183 } 184 } 185 cnt++; 186 } 187 188 sort(shapes, shapes + cnt, cmp); 189 for(int i = 0; i < cnt; i++){ 190 printf("%c ", shapes[i].name); 191 memset(inter, 0, sizeof(inter)); 192 int ans = 0; 193 for(int j = 0; j < cnt; j++){ 194 if(i == j)continue; 195 for(int a = 0; a < shapes[i].nvertices; a++){ 196 for(int b = 0; b < shapes[j].nvertices; b++){ 197 if(intersect_in(shapes[i].p[a], shapes[i].p[(a + 1) % shapes[i].nvertices], shapes[j].p[b], shapes[j].p[(b + 1) % shapes[j].nvertices])){ 198 inter[j] = true; 199 } 200 } 201 } 202 if(inter[j] == true){ 203 ans++; 204 } 205 } 206 if(ans == 0){ 207 printf("has no intersections "); 208 } 209 else{ 210 printf("intersects with"); 211 if(ans == 1){ 212 for(int j = 0; j < cnt; j++){ 213 if(inter[j]){ 214 printf(" %c ", shapes[j].name); 215 } 216 } 217 } 218 else if(ans == 2){ 219 int tmp = 0; 220 for(int j = 0; j < cnt; j++){ 221 if(inter[j]){ 222 if(tmp){ 223 printf(" and"); 224 } 225 printf(" %c", shapes[j].name); 226 tmp++; 227 } 228 } 229 printf(" "); 230 } 231 else{ 232 int tmp = 0; 233 for(int j = 0; j < cnt; j++){ 234 if(inter[j]){ 235 if(tmp == ans - 1){ 236 printf(" and %c", shapes[j].name); 237 } 238 else{ 239 printf(" %c,", shapes[j].name); 240 } 241 tmp++; 242 } 243 244 } 245 printf(" "); 246 } 247 } 248 } 249 printf(" "); 250 } 251 return 0; 252 }