POJ 2653 Pick-up sticks (线段相交)

题意:给你n条线段依次放到二维平面上,问最后有哪些没与前面的线段相交,即它是顶上的线段

 

题解:数据弱,正向纯模拟可过

但是有一个陷阱:如果我们从后面向前枚举,找与前面哪些相交,再删除前面那些相交的线段,这样就错了

因为如果线段8与5,6,7相交了,我们接下来不能直接判断4,我们还要找7,6,5与之前哪些相交

 

#include<set>
#include<map>
#include<queue>
#include<stack>
#include<cmath>
#include<vector>
#include<string>
#include<cstdio>
#include<cstring>
#include<iomanip>
#include<stdlib.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define eps 1E-8
/*注意可能会有输出-0.000*/
#define Sgn(x) (x<-eps? -1 :x<eps? 0:1)//x为两个浮点数差的比较,注意返回整型
#define Cvs(x) (x > 0.0 ? x+eps : x-eps)//浮点数转化
#define zero(x) (((x)>0?(x):-(x))<eps)//判断是否等于0
#define mul(a,b) (a<<b)
#define dir(a,b) (a>>b)
typedef long long ll;
typedef unsigned long long ull;
const int Inf=1<<28;
const ll INF=1ll<<60;
const double Pi=acos(-1.0);
const int Mod=1e9+7;
const int Max=100010;
struct point
{
    double x,y;
};
struct line
{
    point a,b;
};
double xmult(point p1,point p2,point p0)
{
    return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);
}
int isIntersected(point p1,point p2,point l1,point l2)
{
    return (max(p1.x,p2.x)>=min(l1.x,l2.x)) &&
           (max(p1.y,p2.y)>=min(l1.y,l2.y)) &&
           (max(l1.x,l2.x)>=min(p1.x,p2.x)) &&
           (max(l1.y,l2.y)>=min(p1.y,p2.y)) &&
           (xmult(l1,p2,p1)*xmult(p2,l2,p1)>0) &&
           (xmult(p1,l2,l1)*xmult(l2,p2,l1)>0) ;
}
line sti[Max];
int ans[Max],vis[Max];
int Solve(int n)
{
    int coun=0;
    for(int i=1;i<=n;++i)
    {
        for(int j=i+1;j<=n;++j)
        {
            if(isIntersected(sti[i].a,sti[i].b,sti[j].a,sti[j].b))
                {
                    vis[i]=0;
                    break;
                }
        }
    }
    for(int i=n;i;--i)
        if(vis[i])
        ans[coun++]=i;
    return coun;
}
int main()
{
    int n;
    while(~scanf("%d",&n)&&n)
    {
        for(int i=1;i<=n;++i)
        {
            scanf("%lf %lf %lf %lf",&sti[i].a.x,&sti[i].a.y,&sti[i].b.x,&sti[i].b.y);
            vis[i]=1;
        }
        int coun=Solve(n);
        printf("Top sticks: ");
        for(int i=coun-1;~i;--i)
        {
            printf("%d%s",ans[i],i?", ":".\n");
        }
    }
    return 0;
}

 

posted @ 2016-11-05 01:30  专注如一  阅读(245)  评论(0编辑  收藏  举报