POJ 3304 Segments

几何 直线与线段相交

若是每条线段的投影在直线上有重合的点，那么咱们经过这一点作一条直线一定会通过全部的线段！！ 那么咱们考虑把这条直线随意移动到与其中一条线段的某个端点重合，此时直线仍是过了全部线段，咱们再以该点为中心顺时针或逆时针旋转直线，让这条直线刚好通过另外一个线段的某个端点，此时直线必定仍是通过全部线段，且通过了其中某两个线段的两个端点。 根据这个思路咱们能够枚举全部端点，用叉积判断直线与线段是否相交，复杂度O(n^3)

#include <iostream>
#include <cstdio>
#include <cmath>
#define INF 0x3f3f3f3f
#define full(a, b) memset(a, b, sizeof a)
using namespace std;
typedef long long ll;
inline int lowbit(int x){ return x & (-x); }
inline int read(){
    int X = 0, w = 0; char ch = 0;
    while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }
    while(isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
    return w ? -X : X;
}
inline int gcd(int a, int b){ return a % b ? gcd(b, a % b) : b; }
inline int lcm(int a, int b){ return a / gcd(a, b) * b; }
template<typename T>
inline T max(T x, T y, T z){ return max(max(x, y), z); }
template<typename T>
inline T min(T x, T y, T z){ return min(min(x, y), z); }
template<typename A, typename B, typename C>
inline A fpow(A x, B p, C lyd){
    A ans = 1;
    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;
    return ans;
}

const int N = 105;
const double eps = 1e-8;
struct Point { double x,  y;} s[N], e[N];
int n, _;

double mul(const Point &a, const Point &b, const Point &c){
    return (a.x - c.x) * (b.y - c.y) - (a.y - c.y) * (b.x - c.x);
}

bool check(const Point &a, const Point &b){
    if(fabs(a.x - b.x) < eps && fabs(a.y - b.y) < eps) return false;
    for(int i = 0; i < n; i ++){
        if(mul(a, b, s[i]) * mul(a, b, e[i]) > eps)
            return false;
    }
    return true;
}

int main(){

    while(scanf("%d", &_) != EOF){
        for(; _; _ --){
            scanf("%d", &n);
            for(int i = 0; i < n; i ++){
                scanf("%lf%lf%lf%lf", &s[i].x, &s[i].y, &e[i].x, &e[i].y);
            }
            if(n == 1){
                printf("Yes!\n");
                continue;
            }
            bool flag = false;
            for(int i = 0; i < n; i ++){
                for(int j = i + 1; j < n; j ++){
                    if(check(s[i], s[j]) || check(s[i], e[j]) || check(e[i], e[j]) || check(e[i], s[j])){
                        flag = true;
                        break;
                    }
                }
                if(flag) break;
            }
            if(flag) printf("Yes!\n");
            else printf("No!\n");
        }
    }
    return 0;
}