算法题：蚂蚁走木杆

有一根27厘米的细木杆，在第3厘米、7厘米、11厘米、17厘米、23厘米这五个位置上各有一只蚂蚁。 木杆很细，不能同时通过一只蚂蚁。开始 时，蚂蚁的头朝左还是朝右是任意的，它们只会朝前走或调头， 但不会后退。当任意两只蚂蚁碰头时，两只蚂蚁会同时调头朝反方向走。假设蚂蚁们每秒钟可以走一厘米的距离。 编写程序，求所有蚂蚁都离开木杆 的最小时间和最大时间。
http://www.blogjava.net/itspy/archive/2006/11/10/80435.html
下面贴上我的测试程序

namespace ANTS
{
//#define N 5
#define LENGTH 27

    enum STATUS
    {
        Left=0,
        Right=1,
        STOP=2,
    };

    struct ANT 
    {
        double Position;
        STATUS status;
    };

    int goOut(ANT ant)
    {
        if(ant.Position>=0 && ant.Position<=LENGTH)
            return 0;

        if(( ant.Position>=LENGTH && ant.status==Left)||
            ( ant.Position<=0 && ant.status==Right))
        {
            printf("Error:位置%g和状态%s不符合\n",ant.Position,ant.status==Left?"L":"R" );
            exit(0);
        }
        return 1;
    }

    int (__cdecl compare )(const void *e1, const void *e2 )
    {
        ANT *elem1=(ANT *)e1;
        ANT *elem2=(ANT *)e2;
        if(elem1->Position <elem2->Position)
            return -1;
        if(elem1->Position >elem2->Position)
            return 1;
        return 0;
    }
    int Move(ANT *ant,int N,double *timeUsed)
    {
        //how many go out?
        int i,j,k;
        int num=0;
        for (i=0;i<N;i++)
        {
            if (goOut(ant[i]) )
            {
                 ant[i].status=STOP;
                 num++;
            }
        }
        Trace("%d个已经出去的",num);
        if(num>=N)
        {
            Trace("END---------------");
            return 1;
        }

        //
        qsort(ant,N,sizeof(ANT),compare);
        for(i=0;i<N-1;i++)
        {
            if(fabs(ant[i].Position-ant[i+1].Position)<1e-5 && ant[i].status==Right && ant[i+1].status==Left )
            {
                ant[i].status=Left;
                ant[i+1].status=Right;
            }
        }
        Trace("排序后状态");
        for(i=0;i<N;i++)
            Trace("   %g%s ",ant[i].Position,ant[i].status==Left?"L":"R");
        

//         i=0;
//         while (ant[i].status==STOP)
//         {
//             CHECK(i<N);
//             i++;
//         }
// 
//         j=N-1;
//         while (ant[j].status==STOP)
//         {
//             CHECK(j>=0);
//             j--;
//         }
//         
//         CHECK(j-i+1>0);
        //find 最小 碰撞
        int p=-1;
        double len=1e10;
        for (i=0;i<N-1;i++)
        {
            if(ant[i].status==Right && ant[i+1].status==Left 
                && ant[i+1].Position>=0 && ant[i+1].Position<=LENGTH
                && ant[i].Position>=0 && ant[i].Position<=LENGTH
                && ant[i+1].Position-ant[i].Position>1e-5 &&ant[i+1].Position-ant[i].Position<len 
                )
            {
                p=i;
                len=ant[i+1].Position-ant[i].Position;
            }
        }
        if(p>=0)//移动一次直到碰撞
        {
            double lastP1=ant[p].Position;
            double lastP2=ant[p+1].Position;
            double t=len/2;    
            ant[p].Position += t;
            ant[p+1].Position -= t;
            ant[p].status=Left;
            ant[p+1].status=Right;

            for (i=0;i<N;i++)
            {
                if(i==p || i==p+1)
                    continue;
                if(ant[i].status!=STOP)
                    ant[i].Position+=(ant[i].status==Right?1:-1) * t;
            }
            *timeUsed += t;
            Trace("找到最小碰撞的位置%g vs %g,碰撞后时间 %g(dt=%g)",lastP1,lastP2,*timeUsed ,t);
            
            return 0;
        }
        else//所有的都移动
        {
            
            double tMax=0;
            for (i=0;i<N;i++)
            {
                if(ant[i].status!=STOP )
                {
                    if(ant[i].status==Left && ant[i].Position >tMax)
                        tMax=ant[i].Position;
                    else if(ant[i].status==Right && LENGTH - ant[i].Position >tMax)
                        tMax=LENGTH- ant[i].Position;
                }
            }
            *timeUsed+=tMax;
            Trace("没有碰撞,全部出去时间为%g(dt=%g)",*timeUsed,tMax);
            return 1;
        }
    }

    void Test()
    {
        ANT ant[5];
        int Position[5]={3,7,11,17,23};
        int status=0;
        int i;
        double timeUsed=0;

        for(status=0;status<=1/**//*0x1f*/ ;status++)
        {
            printf("初始状态：\n");
            for(i=0;i<5;i++)
            {
                ant[i].Position=Position[i];
                ant[i].status=(status & (1<<i))?Right:Left;
                printf("%g%s ",ant[i].Position,ant[i].status==Left?"L":"R");
            }
            printf("\n");
            
            i=0;
            timeUsed=0;
            while(!Move(ant,5,&timeUsed) && i++<10);
            Trace("END----------");
            printf("使用时间%g\n",timeUsed);
        }
    }
}