poj 2653 Pick-up sticks【线段相交】

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题意：

        依次向地上仍n个棍子（按先后顺序），后仍的如果和前面已经扔过的有交点，那么会叠加上去，

        题目问最后最上面的木棍有哪些。

题解：

        从前向后判断每个木棒上面有没有其他木棒，有的话，直接break。

        否则tle。

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<map>//#include<pair>#include<cmath>#include<vector>#define x first#define y secondusing namespace std;inline int read(){int x=0,f=1;char c=getchar();for(;!isdigit(c);c=getchar())if(c=='-')f=-1;for(;isdigit(c);c=getchar())x=x*10+c-'0';return x*f;}const double eps=1e-8;const double PI=acos(-1.0);const int maxn=1e5+10;int sgn(double x){    if(fabs(x)<eps) return 0;    if(x<0) return -1;    return 1;}struct Point{    double x,y;    Point(){}    Point(double _x,double _y){        x=_x;y=_y;    }    Point operator -(const Point &b)const{        return Point(x-b.x,y-b.y);    }    double operator ^ (const Point &b)const{        return x*b.y - y*b.x;    }};struct Line{    Point s,e;    Line(){}    Line(Point _s,Point _e){        s=_s;e=_e;    }};bool inter(Line l1,Line l2){    return    max(l1.s.x,l1.e.x)>=min(l2.s.x,l2.e.x)&&    max(l2.s.x,l2.e.x)>=min(l1.s.x,l1.e.x)&&    max(l1.s.y,l1.e.y)>=min(l2.s.y,l2.e.y)&&    max(l2.s.y,l2.e.y)>=min(l1.s.y,l1.e.y)&&    sgn((l2.s-l1.e)^(l1.s-l1.e))*sgn((l2.e-l1.e)^(l1.s-l1.e))<=0&&    sgn((l1.s-l2.e)^(l2.s-l2.e))*sgn((l1.e-l2.e)^(l2.s-l2.e))<=0;}Point p1,p2,p3,p4;Line l[maxn],t1;bool v[maxn];vector<int>vec;int main(){    int n,cnt;    double x1,x2,y2,y1;    while(scanf("%d",&n)&&n){        memset(v,0,sizeof(v));        vec.clear();        for(int i=1;i<=n;++i){            scanf("%lf %lf %lf %lf",&x1,&y1,&x2,&y2);            //x1=read();y1=read();x2=read();y2=read();            p1={x1,y1};p2={x2,y2};            l[i]={p1,p2};        }        for(int i=1;i<=n;++i){            for(int j=i+1;j<=n;++j){                if(inter(l[i],l[j])){                    v[i]=true;                    break;                }            }        }        printf("Top sticks: ");        int f=0;        for(int i=1;i<=n;++i)            if(!v[i]){                if(!f)printf("%d",i),f=1;                else printf(", %d",i);            }        puts(".");    }return 0;}