POJ 2653 Pick-up sticks 计算几何 线段相交

计算几何的题目   庆幸不用判断共线  题目说了  

 第一种方法是读入一个处理一个   结果超时了

第二种方法是读完了后再处理    AC    

同样是暴力求解  但是也得讲究技术啊

TLE代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define Max 100005
#define eps 1e-10
using namespace std;

int v[Max];//标记每层的有多少木棍

typedef struct
{
    double x,y;
}point;

typedef struct
{
    point a,b;
    int n;//标记在第几层
}line;

line  stick[Max];

double multi(point p0, point p1, point p2)
{
    return ( p1.x - p0.x )*( p2.y - p0.y )-( p2.x - p0.x )*( p1.y - p0.y );
}

bool isIntersected(point s1,point e1, point s2,point e2)
{
    return  (max(s1.x,e1.x) >= min(s2.x,e2.x))  &&
            (max(s2.x,e2.x) >= min(s1.x,e1.x))  &&
            (max(s1.y,e1.y) >= min(s2.y,e2.y))  &&
            (max(s2.y,e2.y) >= min(s1.y,e1.y))  &&
            (multi(s1,s2,e1)*multi(s1,e1,e2)>0) &&
            (multi(s2,s1,e2)*multi(s2,e2,e1)>0);
}

int main()
{
    #ifdef LOCAL
    freopen("in.txt","r",stdin);
    #endif // LOCAL
    int N;
    while((scanf("%d",&N),N))
    {
        int Maxn  = 1;
        memset(v,0,sizeof(v));
        for(int i = 0; i < N; i++)//读入一个处理一个
        {
            scanf("%lf%lf%lf%lf",&stick[i].a.x,&stick[i].a.y,&stick[i].b.x,&stick[i].b.y);
            stick[i].n = 1;
            for(int j = i-1; j >= 0; j--)
            {
                if(isIntersected(stick[i].a,stick[i].b,stick[j].a,stick[j].b))
                    stick[i].n = stick[j].n + 1;//相交在它的上面一层
                if(stick[i].n > Maxn) {Maxn++;break;}
            }
            v[stick[i].n]++;
        }
        printf("Top sticks:");
        int t = 1;
        for(int i = 0; i < N&&t <=v[Maxn]; i++)//最顶层的木棒数
            if(stick[i].n == Maxn)
            {
                if(t < v[Maxn]){printf(" %d,",i+1);t++;}
                else {printf(" %d.\n",i+1);t++;}
            }
    }
    return 0;
}

下面是AC代码

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define Max 100005
#define eps 1e-10
using namespace std;

int v[Max];//标记每层的有多少木棍

typedef struct
{
    double x,y;
}point;

typedef struct
{
    point a,b;
}line;

line  stick[Max];

double multi(point p0, point p1, point p2)
{
    return ( p1.x - p0.x )*( p2.y - p0.y )-( p2.x - p0.x )*( p1.y - p0.y );
}

bool isIntersected(point s1,point e1, point s2,point e2)
{
    return  (max(s1.x,e1.x) >= min(s2.x,e2.x))  &&
            (max(s2.x,e2.x) >= min(s1.x,e1.x))  &&
            (max(s1.y,e1.y) >= min(s2.y,e2.y))  &&
            (max(s2.y,e2.y) >= min(s1.y,e1.y))  &&
            (multi(s1,s2,e1)*multi(s1,e1,e2)>0) &&
            (multi(s2,s1,e2)*multi(s2,e2,e1)>0);
}

int main()
{
    #ifdef LOCAL
    freopen("in.txt","r",stdin);
    #endif // LOCAL
    int N;
    while((scanf("%d",&N),N))
    {
        for(int i = 0;i < N; i++)v[i] = 0;
        for(int i = 0; i < N; i++)//读入一个处理一个
            scanf("%lf%lf%lf%lf",&stick[i].a.x,&stick[i].a.y,&stick[i].b.x,&stick[i].b.y);
        for(int i = 0; i < N; i++) //  从头扫到尾  只需一遍判断
            for(int j = i+1; j < N; j++)
        {
            if(isIntersected(stick[i].a,stick[i].b,stick[j].a,stick[j].b))
            {v[i] = 1;break;}//标记不是在最上面一层
        }
        printf("Top sticks:");
        int t = N-1;
        while(v[t]) t--;
        for(int i = 0; i <= t; i++)
            if(v[i]== 0)
            {
                if(i < t)printf(" %d,",i+1);//最顶层的木棒数
                else printf(" %d.\n",i+1);
            }
    }
    return 0;
}