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POJ1363：Rails

2013-07-22 14:59 351 查看
Description
ThereisafamousrailwaystationinPopPushCity.Countrythereisincrediblyhilly.Thestationwasbuiltinlastcentury.Unfortunately,fundswereextremelylimitedthattime.Itwaspossibletoestablishonlyasurfacetrack.
Moreover,itturnedoutthatthestationcouldbeonlyadead-endone(seepicture)andduetolackofavailablespaceitcouldhaveonlyonetrack.



ThelocaltraditionisthateverytrainarrivingfromthedirectionAcontinuesinthedirectionBwithcoachesreorganizedinsomeway.AssumethatthetrainarrivingfromthedirectionAhasN<=1000coachesnumberedinincreasingorder1,2,...,N.The
chieffortrainreorganizationsmustknowwhetheritispossibletomarshalcoachescontinuinginthedirectionBsothattheirorderwillbea1,a2,...,aN.Helphimandwriteaprogramthatdecideswhetheritispossibletogettherequiredorderofcoaches.
YoucanassumethatsinglecoachescanbedisconnectedfromthetrainbeforetheyenterthestationandthattheycanmovethemselvesuntiltheyareonthetrackinthedirectionB.Youcanalsosupposethatatanytimetherecanbelocatedasmanycoaches
asnecessaryinthestation.ButonceacoachhasenteredthestationitcannotreturntothetrackinthedirectionAandalsoonceithasleftthestationinthedirectionBitcannotreturnbacktothestation.

Input
Theinputconsistsofblocksoflines.Eachblockexceptthelastdescribesonetrainandpossiblymorerequirementsforitsreorganization.InthefirstlineoftheblockthereistheintegerNdescribedabove.Ineachofthe
nextlinesoftheblockthereisapermutationof1,2,...,N.Thelastlineoftheblockcontainsjust0.

Thelastblockconsistsofjustonelinecontaining0.
Output
Theoutputcontainsthelinescorrespondingtothelineswithpermutationsintheinput.AlineoftheoutputcontainsYesifitispossibletomarshalthecoachesintheorderrequiredonthecorrespondinglineoftheinput.Otherwise
itcontainsNo.Inaddition,thereisoneemptylineafterthelinescorrespondingtooneblockoftheinput.Thereisnolineintheoutputcorrespondingtothelast``null''blockoftheinput.
SampleInput
5
12345
54123
0
6
654321
0
0

SampleOutput
Yes
No

Yes


题意：按1~n进栈，然后给出一个顺序，看能否按照这个顺序出栈


#include<stdio.h>

intmain()
{
inta[1005],b[1005],i,j,k,n;
while(scanf("%d",&n),n)
{
while(scanf("%d",&b[0]),b[0])
{
for(j=1;j<n;j++)
scanf("%d",&b[j]);
for(i=1,j=0,k=0;i<=n&&j<n;i++,k++)
{
a[k]=i;
while(a[k]==b[j])
{
if(k>0)k--;
else
{
a[k]=0,k--;
}
j++;
if(k==-1)break;
}
}
if(j==n)printf("Yes\n");
elseprintf("No\n");
}
printf("\n");
}
}

                                            

                                        

                        
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