poj3304 Segments

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多组数据:给n条线段 判断是否有一条直线能与所有的线段产生交♂点

首先n=1/n=2显然

然后考虑如果有一种方案是从中间穿过去 那么一定有某种方案

使得这条直线绕某个点旋转某个角度

然后交上一个线段的端点

所以枚举端点对然后去交线段就行

注意有一个交不上就验证下一个 全都不成立再输出no等等

Code:

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<iostream>
#define ms(a,b) memset(a,b,sizeof a)
#define rep(i,a,n) for(int i = a;i <= n;i++)
#define per(i,n,a) for(int i = n;i >= a;i--)
#define inf 2147483647
using namespace std;
typedef long long ll;
typedef double D;
#define eps 1e-8
ll read() {
    ll as = 0,fu = 1;
    char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') fu = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        as = as * 10 + c - '0';
        c = getchar();
    }
    return as * fu;
}
//head
#define P point
struct point {
    D x,y;
    point(){}
    point(D X,D Y):x(X),y(Y){}
    D len() {return sqrt(x*x+y*y);}
    D tan() {return y/x;}
    D sin() {return len() / y;}
    D cos() {return len() / x;}
    void read() {scanf("%lf%lf",&x,&y);}
    void print() {printf("%.2lf %.2lf
",x,y);}
    friend inline point operator + (const point &a,const point &b) {
        return point(a.x+b.x,a.y+b.y);
    }
    friend inline point operator - (const point &a,const point &b) {
        return point(a.x-b.x,a.y-b.y);
    }
    //放缩
    friend inline point operator * (const point &a,const D &b) {
        return point(a.x*b,a.y*b);
    }
    //叉积
    friend inline D operator * (const point &a,const point &b) {
        return a.x*b.y-a.y*b.x;
    }
    //点积
    friend inline D operator / (const point &a,const point &b) {
        return a.x*b.x+a.y*b.y;
    }
    bool operator == (const point &o) const {
        return (x == o.x) && (y == o.y);
    }
};
struct yuan {
    D r,x,y;
    yuan(){}
    yuan(int R,int X,int Y):r(R),x(X),y(Y){}
};
struct line {
    point x,y;
};
bool ONSEG(point a,point b,point p) {
    return ((a-b).len() == (a-p).len() + (p-b).len());
}
D TRIAREA(point a,point b,point c) {
    return ((a-b)*(a-c)) / 2.0;
}
#define ONLINE(a,b,c) (((a-b)*(a-c)) == 0)
#define sign(x) (x) > 0 ? 1 : ((x) < 0 ? -1 : 0)
// 1  0 -1
// 锐 直 钝
int ANGDIR(point a,point b,point p) {
    D ans = (p-a)*(p-b);
    return sign(ans);
}
D dis(point a,point b,point p) {
    if(ANGDIR(b,p,a) == -1) return (p-a).len();
    if(ANGDIR(a,p,b) == -1) return (p-b).len();
    return ((p-a)*(p-b)) / (a-b).len();
}
int cross(P a,P b,P c,P d) {
    if(ONLINE(a,b,c) ^ ONLINE(a,b,d)) return 1;
    if(ONLINE(c,d,a) ^ ONLINE(c,d,b)) return 1;
    if(ONLINE(a,b,c) & ONLINE(a,b,d)) return -1;
    if(ONLINE(c,d,a) & ONLINE(c,d,b)) return -1;
    D J1 = ((c-d)*(c-a)) * ((c-d)*(c-b));
    D J2 = ((a-b)*(a-c)) * ((a-b)*(a-d));
    if(J1 < 0 && J2 < 0) return 1;
    return 0;
}
point Cross(point a,point b,point c,point d) {
    if(ONLINE(a,b,c)) return c;
    if(ONLINE(a,b,d)) return d;
    if(ONLINE(c,d,a)) return a;
    if(ONLINE(c,d,b)) return b;
    D S1 = (a-c)*(a-d),S2 = (b-d) * (b-c);
    point tmp = (b-a) * (S1 / (S1+S2));
    return a + tmp;
}
//CP
int n;
const int N = 2006;
point p[N<<1];

bool check(int x,int y) {
    point a,b;
    rep(k,1,n) {
        a = p[k-1<<1|1],b = p[k<<1];
        if(((a-p[x])*(a-p[y])) * ((b-p[x])*(b-p[y])) > 0.0) return 0;
    }
    return 1;
}

bool solve() {
    n = read();
    rep(i,1,n<<1) scanf("%lf%lf",&p[i].x,&p[i].y);
    rep(i,1,n<<1) rep(j,i+1,n<<1) {
        if(p[i] == p[j]) continue;
        if(check(i,j)) return 1;
    }
    return 0;
}

int main() {
    int T = read();
    while(T--) solve() ? puts("Yes!") : puts("No!");
    return 0;
}

依然一大堆函数没用qwq