BZOJ2494 Triangles and Quadrangle

一道非常"简单"的计算几何题。。。

题意：给你两个三角形和一个四边形，问你能否用这两个三角形拼成这个四边形

首先。。。四边形可能是凹四边形。。。需要判断一下。。。这个比较简单直接分成两部分即可。。。

然后。。。四边形可能会退化成三角形。。。那就需要两个三角形拼起来的时候某两个角之和为180°

最后暴力一下如何划分四边形即可。。。

写的真是开心，还好一遍A了QAQ

 

  1 /**************************************************************
  2     Problem: 2494
  3     User: rausen
  4     Language: C++
  5     Result: Accepted
  6     Time:8 ms
  7     Memory:820 kb
  8 ****************************************************************/
  9  
 10 #include <cstdio>
 11 #include <cmath>
 12 #include <algorithm>
 13  
 14 using namespace std;
 15 typedef double lf;
 16  
 17 template <class T> inline T sqr(const T &x) {
 18     return x * x;
 19 }
 20  
 21 template <class T> inline int sgn(const T &x) {
 22     const T eps = 1e-8;
 23     if (fabs(x) < eps) return 0;
 24     return x < 0 ? -1 : 1;
 25 }
 26  
 27 struct point {
 28     lf x, y;
 29     point() {}
 30     point(lf _x, lf _y) : x(_x), y(_y) {}
 31  
 32     inline point operator + (const point &p) const {
 33         return point(x + p.x, y + p.y);
 34     }
 35     inline point operator - (const point &p) const {
 36         return point(x - p.x, y - p.y);
 37     }
 38     inline lf operator * (const point &p) const {
 39         return x * p.y - y * p.x;
 40     }
 41     inline point operator * (const lf &d) const {
 42         return point(x * d, y * d);
 43     }
 44     inline void get() {
 45         scanf("%lf%lf", &x, &y);
 46     }
 47     friend inline lf dis2(const point &p) {
 48         return sqr(p.x) + sqr(p.y);
 49     }
 50     friend inline lf dis(const point &p) {
 51         return sqrt(dis2(p));
 52     }
 53 } t[2][6], p[8];
 54  
 55 inline bool check(const point &a, const point &b, const point &c, const int &p) {
 56     static lf T1[3], T2[3];
 57     static int i;
 58     T1[0] = dis(a - b), T1[1] = dis(b - c), T1[2] = dis(c - a);
 59     for (i = 0; i < 3; ++i)
 60         T2[i] = dis(t[p][i] - t[p][i + 1]);
 61     sort(T1, T1 + 3), sort(T2, T2 + 3);
 62     for (i = 0; i < 3; ++i)
 63         if (sgn(T1[i] - T2[i]) != 0) return 0;
 64     return 1;
 65 }
 66  
 67 inline point get_cross(const point &a, const point &b, const point &c, const point &d) {
 68     static lf rate;
 69     rate = 1 / (1 - ((d - c) * (b - c)) / ((d - c) * (a - c)));
 70     return (b - a) * rate + a;
 71 }
 72  
 73 int main() {
 74     int T, icase, i, j, f;
 75     lf di[6], d1;
 76     point p1, a, b, c, d, e;
 77     scanf("%d", &T);
 78     for (icase = 1; icase <= T; ++icase) {
 79         f = 0;
 80         for (i = 0; i < 3; ++i) t[0][i].get(), t[0][i + 3] = t[0][i];
 81         for (i = 0; i < 3; ++i) t[1][i].get(), t[1][i + 3] = t[1][i];
 82         for (i = 0; i < 4; ++i) p[i].get(), p[i + 4] = p[i];
 83          
 84         for (i = 0; i < 3; ++i)
 85             di[i] = dis(t[0][i] - t[0][i + 1]), di[i + 3] = dis(t[1][i] - t[1][i + 1]);
 86         if (((p[2] - p[0]) * (p[1] - p[0])) * ((p[2] - p[0]) * (p[3] - p[0])) < 0) {
 87             if (check(p[0], p[1], p[2], 0) && check(p[0], p[2], p[3], 1)) f = 1;
 88             if (check(p[0], p[1], p[2], 1) && check(p[0], p[2], p[3], 0)) f = 1;
 89         }
 90         if (((p[3] - p[1]) * (p[0] - p[1])) * ((p[3] - p[1]) * (p[2] - p[1])) < 0) {
 91             if (check(p[1], p[3], p[0], 0) && check(p[1], p[3], p[2], 1)) f = 1;
 92             if (check(p[1], p[3], p[0], 1) && check(p[1], p[3], p[2], 0)) f = 1;
 93         }
 94          
 95         for (i = 0; i < 4; ++i) {
 96             a = p[i], b = p[i + 1], c = p[i + 2], d = p[i + 3];
 97             if (((b - a) * (c - a)) * ((b - a) * (d - a)) < 0) {
 98                 e = get_cross(a, b, c, d);
 99                 if (check(a, d, e, 0) && check(b, c, e, 1)) f = 1;
100                 if (check(a, d, e, 1) && check(b, c, e, 0)) f = 1;
101             }
102         }
103         for (i = 0; i < 4; ++i) if (sgn((p[i + 1] - p[i]) * (p[i + 2] - p[i])) == 0) {
104             d1 = dis(p[i + 2] - p[i]);
105             for (j = 0; j < 6; ++j) if (d1 >= di[j]) {
106                 p1 = p[i + 2] - p[i];
107                 p1 = p1 * (di[j] / d1) + p[i];
108                 if (check(p[i + 3], p[i], p1, 0) && check(p[i + 3], p[i + 2], p1, 1)) f = 1;
109                     if (check(p[i + 3], p[i], p1, 1) && check(p[i + 3], p[i + 2], p1, 0)) f = 1;
110             }
111         }
112         printf("Case #%d: %s\n", icase, f ? "Yes" : "No");
113     }
114     return 0;
115 }
View Code

 

posted on 2015-03-24 22:36  Xs酱~  阅读(308)  评论(0)  编辑  收藏  举报