POJ 2653 Pick-up sticks

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链接点击：POJ 2653 Pick-up sticks

一道依靠纯粹暴力就可以过得简单题。

#include<stdio.h>#include<string.h>#include<math.h>#define eps 1e-6const int MAXN=100002;bool vis[MAXN];typedef struct{    double x,y;}Point;Point p1[MAXN],p2[MAXN];int dblcmp(double x){    if(fabs(x)<eps)      return 0;    return x>0?1:-1;}double det(double x1,double y1,double x2,double y2){  //求叉积    return x1*y2-x2*y1;}double cross(Point a,Point b,Point c){    return det(a.x-c.x,a.y-c.y,b.x-c.x,b.y-c.y);}bool segcross(Point a,Point b,Point c,Point d){    int d1,d2,d3,d4;    d1=dblcmp(cross(a,b,c));    d2=dblcmp(cross(a,b,d));    d3=dblcmp(cross(c,d,a));    d4=dblcmp(cross(c,d,b));    if((d1^d2)==-2&&(d3^d4)==-2)      return true;    return false;}int main(){    int n,i,j;    while(scanf("%d",&n),n)    {        memset(vis,true,sizeof(vis));        for(i=1;i<=n;i++)          scanf("%lf%lf%lf%lf",&p1[i].x,&p1[i].y,&p2[i].x,&p2[i].y);        for(i=1;i<=n;i++)          for(j=n;j>=i+1;j--)//只能倒着遍历，否则会超时，因为后面压前面的概率大，所以从后开始找更加省时。            if(segcross(p1[i],p2[i],p1[j],p2[j]))//判断线段是否相交            {                vis[i]=false;                break;            }        printf("Top sticks: ");        for(i=1;i<n;i++)          if(vis[i])            printf("%d, ",i);        printf("%d.\n",n);    }    return 0;}