題意:n個線段,一個一個的放,問最後沒有被壓住的都有哪些(貌似不用考慮端點?題中沒細說)
思路:n有點大,可以考慮用兩個隊列(不過比直接算慢了點,因爲用了stl?),或者是直接算,如果沒有和後面的相交就可以,否則跳過
代碼:
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<cmath>
#include<queue>
#include<list>
using namespace std;
const double INF = 1e200;
const double EP = 1e-8;
const int maxn = 100000+100;
const double PI = acos(-1);
struct POINT{///點 定義
double x;
double y;
POINT(double a = 0,double b = 0){x = a;y = b;}
};
struct SEGMENT{///line segment///線段 定義
POINT s;
POINT e;
SEGMENT(POINT a,POINT b){s = a;e = b;}
SEGMENT(){}
};
struct LINE{///ax + by + c = 0&&a >= 0///一般式
double a;
double b;
double c;
LINE(double da,double db,double dc){a = da;b = db;c = dc;}
LINE(double x1,double y1,double x2,double y2){///根據兩個點求出一般式
a = y1 - y2;b = x2 - x1;c = x1*y2 - x2*y1;
if(a < 0){a*=-1;b*=-1;c*=-1;}
}
};
double multiply(POINT sp,POINT ep,POINT op){///向量op->sp X op->ep的叉乘,小於0:ep在op->sp順時針方向//大於0:0:ep在op->sp逆時針方向//等於0:三點共線
return ((sp.x - op.x)*(ep.y - op.y) - (ep.x - op.x)*(sp.y - op.y));
}
bool online(SEGMENT l,POINT p){///判斷點是否在線段上
return ((multiply(l.e,p,l.s) == 0)&&(((p.x-l.s.x)*(p.x-l.e.x))<= 0)&&(((p.y-l.s.y)*(p.y-l.e.y)) <= 0));
}
bool intersect(SEGMENT u,SEGMENT v){///兩線段相交(包括端點),返回true
return ((max(u.s.x,u.e.x) >= min(v.s.x,v.e.x))&&
(max(v.s.x,v.e.x) >= min(u.s.x,u.e.x))&&
(max(u.s.y,u.e.y) >= min(v.s.y,v.e.y))&&
(max(v.s.y,v.e.y) >= min(u.s.y,u.e.y))&&
(multiply(v.s,u.e,u.s)*multiply(u.e,v.e,u.s) >= 0)&&
(multiply(u.s,v.e,v.s)*multiply(v.e,u.e,v.s) >= 0));
}
bool intersect_a(SEGMENT u,SEGMENT v){///兩線段相交(不包括端點)
return ((intersect(u,v))&&
!online(u,v.s)&&
!online(u,v.e)&&
!online(v,u.e)&&
!online(v,u.s));
}
int lineintersect(LINE l1,LINE l2,POINT &p){///求兩直線交點,有交點返回1和交點,沒有返回0,重合返回2
double d = l1.a*l2.b-l2.a*l1.b;
double d2 = l1.a*l2.c-l2.a*l1.c;
double d3 = l1.b*l2.c-l2.b*l1.c;
if(fabs(d) < EP&&fabs(d2) < EP&&fabs(d3) < EP)return 2;
p.x = (l2.c*l1.b-l1.c*l2.b)/d;
p.y = (l2.a*l1.c-l1.a*l2.c)/d;
if(fabs(d) < EP)return 0;
return 1;
}
double point_dis(POINT a,POINT b){///求兩點距離
return sqrt((a.x - b.x)*(a.x - b.x) + (a.y - b.y)*(a.y - b.y));
}
///************************************************************************************************************************************
int n;
SEGMENT ss[maxn];
int main(){
//freopen("in.txt","r",stdin);
while(~scanf("%d",&n)){
if(n == 0)break;
for(int i = 0;i < n;i ++){
double a,b,c,d;
scanf("%lf%lf%lf%lf",&a,&b,&c,&d);
POINT aa(a,b),bb(c,d);
ss[i].s = aa;ss[i].e = bb;
}
printf("Top sticks:");
for(int i = 0;i < n-1;i++){
bool flag = 1;
for(int j = i+1;j < n;j ++){
if(intersect(ss[i],ss[j])){
flag = 0;
break;
}
}
if(flag)printf(" %d,",i+1);
}
printf(" %d.\n",n);
}
return 0;
}