poj3304 判断是否存在一条直线经过n条线段

枚举两条线段的端点即可

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<string>
#include<stdlib.h>
#include<cmath>
#include<algorithm>
using namespace std;
#define rd(x) scanf("%d",&x)
#define rdd(x,y) scanf("%d%d",&x,&y)
#define rddd(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define rds(s) scanf("%s",s)
#define rep(i,n) for(int i=0;i<n;i++)
#define LL long long
const int N = 1e2+10;
const int M=5e5+10;
const int inf=1e9;
const double esp=1e-8;
const int MOD=1e9+7;
int n,m;
struct Point{
    double x,y;
    Point(){

    }
    Point(double _x,double _y){
        x=_x;y=_y;
    }
};
struct Seg{
    Point s,e;
    Seg(){}
    Seg(Point _s,Point _e){
        s=_s;
        e=_e;
    }
}seg[N];
double cross(Point a,Point b,Point c){
    return (b.x-a.x)*(c.y-a.y)-(b.y-a.y)*(c.x-a.x);
}
double dis(Point a,Point b){
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
int dcmp(double x){
    if(fabs(x)<esp) return 0;
    return x<0?-1:1;
}
bool judge(Point a,Point b){
    if(dcmp(dis(a,b)==0) ) return 0;
    for(int i=1;i<=n;i++){
        int ret=dcmp(cross(a,b,seg[i].s))*dcmp(cross(a,b,seg[i].e));
        if(ret>0) return false;
    }
    return true;
}
bool solve(){
    if(n<=2) return true;
     for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++) if(i!=j){
                if(judge(seg[i].s,seg[j].s)) return true;
                if(judge(seg[i].s,seg[j].e)) return true;
                if(judge(seg[i].e,seg[j].s)) return true;
                if(judge(seg[i].e,seg[j].e)) return true;
        }
    return false;
}
int main()
{
#ifndef ONLINE_JUDGE
   freopen("aaa","r",stdin);
#endif
    int T;
    rd(T);
    while(T--){
        rd(n);
        for(int i=1;i<=n;i++){
            double u,v,w,z;
            scanf("%lf%lf%lf%lf",&u,&v,&w,&z);
            seg[i]=Seg(Point(u,v),Point(w,z));
        }
        printf("%s\n",solve()?"Yes!":"No!");

    }
    return 0;
}


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