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

 /**************************************************************
     Problem: 2494
     User: rausen
     Language: C++
     Result: Accepted
     Time:8 ms
     Memory:820 kb
 ****************************************************************/

 #include <cstdio>
 #include <cmath>
 #include <algorithm>

 using namespace std;
 typedef double lf;

 template <class T> inline T sqr(const T &x) {
     return x * x;
 }

 template <class T> inline int sgn(const T &x) {
     const T eps = 1e-;
     if (fabs(x) < eps) return ;
     return x <  ? - : ;
 }

 struct point {
     lf x, y;
     point() {}
     point(lf _x, lf _y) : x(_x), y(_y) {}

     inline point operator + (const point &p) const {
         return point(x + p.x, y + p.y);
     }
     inline point operator - (const point &p) const {
         return point(x - p.x, y - p.y);
     }
     inline lf operator * (const point &p) const {
         return x * p.y - y * p.x;
     }
     inline point operator * (const lf &d) const {
         return point(x * d, y * d);
     }
     inline void get() {
         scanf("%lf%lf", &x, &y);
     }
     friend inline lf dis2(const point &p) {
         return sqr(p.x) + sqr(p.y);
     }
     friend inline lf dis(const point &p) {
         return sqrt(dis2(p));
     }
 } t[][], p[];

 inline bool check(const point &a, const point &b, const point &c, const int &p) {
     static lf T1[], T2[];
     static int i;
     T1[] = dis(a - b), T1[] = dis(b - c), T1[] = dis(c - a);
     for (i = ; i < ; ++i)
         T2[i] = dis(t[p][i] - t[p][i + ]);
     sort(T1, T1 + ), sort(T2, T2 + );
     for (i = ; i < ; ++i)
         if (sgn(T1[i] - T2[i]) != ) return ;
     return ;
 }

 inline point get_cross(const point &a, const point &b, const point &c, const point &d) {
     static lf rate;
     rate =  / ( - ((d - c) * (b - c)) / ((d - c) * (a - c)));
     return (b - a) * rate + a;
 }

 int main() {
     int T, icase, i, j, f;
     lf di[], d1;
     point p1, a, b, c, d, e;
     scanf("%d", &T);
     for (icase = ; icase <= T; ++icase) {
         f = ;
         for (i = ; i < ; ++i) t[][i].get(), t[][i + ] = t[][i];
         for (i = ; i < ; ++i) t[][i].get(), t[][i + ] = t[][i];
         for (i = ; i < ; ++i) p[i].get(), p[i + ] = p[i];

         for (i = ; i < ; ++i)
             di[i] = dis(t[][i] - t[][i + ]), di[i + ] = dis(t[][i] - t[][i + ]);
         if (((p[] - p[]) * (p[] - p[])) * ((p[] - p[]) * (p[] - p[])) < ) {
             if (check(p[], p[], p[], ) && check(p[], p[], p[], )) f = ;
             if (check(p[], p[], p[], ) && check(p[], p[], p[], )) f = ;
         }
         if (((p[] - p[]) * (p[] - p[])) * ((p[] - p[]) * (p[] - p[])) < ) {
             if (check(p[], p[], p[], ) && check(p[], p[], p[], )) f = ;
             if (check(p[], p[], p[], ) && check(p[], p[], p[], )) f = ;
         }

         for (i = ; i < ; ++i) {
             a = p[i], b = p[i + ], c = p[i + ], d = p[i + ];
             if (((b - a) * (c - a)) * ((b - a) * (d - a)) < ) {
                 e = get_cross(a, b, c, d);
                 if (check(a, d, e, ) && check(b, c, e, )) f = ;
                 if (check(a, d, e, ) && check(b, c, e, )) f = ;
             }
         }
         for (i = ; i < ; ++i) if (sgn((p[i + ] - p[i]) * (p[i + ] - p[i])) == ) {
             d1 = dis(p[i + ] - p[i]);
             for (j = ; j < ; ++j) if (d1 >= di[j]) {
                 p1 = p[i + ] - p[i];
                 p1 = p1 * (di[j] / d1) + p[i];
                 if (check(p[i + ], p[i], p1, ) && check(p[i + ], p[i + ], p1, )) f = ;
                     if (check(p[i + ], p[i], p1, ) && check(p[i + ], p[i + ], p1, )) f = ;
             }
         }
         printf("Case #%d: %s\n", icase, f ? "Yes" : "No");
     }
     return ;
 }