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
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enum STATUS
{
Left=0,
Right=1,
STOP=2,
};
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struct ANT
{
double Position;
STATUS status;
};
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int goOut(ANT ant)
{
if(ant.Position>=0 && ant.Position<=LENGTH)
return 0;
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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;
}
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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;
}
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//
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");
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// 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;
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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;
}
}
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void Test()
{
ANT ant[5];
int Position[5]={3,7,11,17,23};
int status=0;
int i;
double timeUsed=0;
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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);
}
}
}
http://www.blogjava.net/itspy/archive/2006/11/10/80435.html
下面贴上我的测试程序
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