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-8;
    if (fabs(x) < eps) return 0;
    return x < 0 ? -1 : 1;
}
 
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[2][6], p[8];
 
inline bool check(const point &a, const point &b, const point &c, const int &p) {
    static lf T1[3], T2[3];
    static int i;
    T1[0] = dis(a - b), T1[1] = dis(b - c), T1[2] = dis(c - a);
    for (i = 0; i < 3; ++i)
        T2[i] = dis(t[p][i] - t[p][i + 1]);
    sort(T1, T1 + 3), sort(T2, T2 + 3);
    for (i = 0; i < 3; ++i)
        if (sgn(T1[i] - T2[i]) != 0) return 0;
    return 1;
}
 
inline point get_cross(const point &a, const point &b, const point &c, const point &d) {
    static lf rate;
    rate = 1 / (1 - ((d - c) * (b - c)) / ((d - c) * (a - c)));
    return (b - a) * rate + a;
}
 
int main() {
    int T, icase, i, j, f;
    lf di[6], d1;
    point p1, a, b, c, d, e;
    scanf("%d", &T);
    for (icase = 1; icase <= T; ++icase) {
        f = 0;
        for (i = 0; i < 3; ++i) t[0][i].get(), t[0][i + 3] = t[0][i];
        for (i = 0; i < 3; ++i) t[1][i].get(), t[1][i + 3] = t[1][i];
        for (i = 0; i < 4; ++i) p[i].get(), p[i + 4] = p[i];
         
        for (i = 0; i < 3; ++i)
            di[i] = dis(t[0][i] - t[0][i + 1]), di[i + 3] = dis(t[1][i] - t[1][i + 1]);
        if (((p[2] - p[0]) * (p[1] - p[0])) * ((p[2] - p[0]) * (p[3] - p[0])) < 0) {
            if (check(p[0], p[1], p[2], 0) && check(p[0], p[2], p[3], 1)) f = 1;
            if (check(p[0], p[1], p[2], 1) && check(p[0], p[2], p[3], 0)) f = 1;
        }
        if (((p[3] - p[1]) * (p[0] - p[1])) * ((p[3] - p[1]) * (p[2] - p[1])) < 0) {
            if (check(p[1], p[3], p[0], 0) && check(p[1], p[3], p[2], 1)) f = 1;
            if (check(p[1], p[3], p[0], 1) && check(p[1], p[3], p[2], 0)) f = 1;
        }
         
        for (i = 0; i < 4; ++i) {
            a = p[i], b = p[i + 1], c = p[i + 2], d = p[i + 3];
            if (((b - a) * (c - a)) * ((b - a) * (d - a)) < 0) {
                e = get_cross(a, b, c, d);
                if (check(a, d, e, 0) && check(b, c, e, 1)) f = 1;
                if (check(a, d, e, 1) && check(b, c, e, 0)) f = 1;
            }
        }
        for (i = 0; i < 4; ++i) if (sgn((p[i + 1] - p[i]) * (p[i + 2] - p[i])) == 0) {
            d1 = dis(p[i + 2] - p[i]);
            for (j = 0; j < 6; ++j) if (d1 >= di[j]) {
                p1 = p[i + 2] - p[i];
                p1 = p1 * (di[j] / d1) + p[i];
                if (check(p[i + 3], p[i], p1, 0) && check(p[i + 3], p[i + 2], p1, 1)) f = 1;
                    if (check(p[i + 3], p[i], p1, 1) && check(p[i + 3], p[i + 2], p1, 0)) f = 1;
            }
        }
        printf("Case #%d: %s
", icase, f ? "Yes" : "No");
    }
    return 0;
}