解决acm上的wa的问题：栈输出序列
在北大acm online上做题，1363题 ，总是wrong answer。现附上题目，和我的解答，请指教！
Rails
Time Limit:1000MS Memory Limit:10000K
Total Submit:5848 Accepted:2297
Description
There is a famous railway station in PopPush City. Country there is incredibly hilly. The station was built in last century. Unfortunately, funds were extremely limited that time. It was possible to establish only a surface track. Moreover, it turned out that the station could be only a dead-end one (see picture) and due to lack of available space it could have only one track.
The local tradition is that every train arriving from the direction A continues in the direction B with coaches reorganized in some way. Assume that the train arriving from the direction A has N <= 1000 coaches numbered in increasing order 1, 2, ..., N. The chief for train reorganizations must know whether it is possible to marshal coaches continuing in the direction B so that their order will be a1, a2, ..., aN. Help him and write a program that decides whether it is possible to get the required order of coaches. You can assume that single coaches can be disconnected from the train before they enter the station and that they can move themselves until they are on the track in the direction B. You can also suppose that at any time there can be located as many coaches as necessary in the station. But once a coach has entered the station it cannot return to the track in the direction A and also once it has left the station in the direction B it cannot return back to the station.
Input
The input consists of blocks of lines. Each block except the last describes one train and possibly more requirements for its reorganization. In the first line of the block there is the integer N described above. In each of the next lines of the block there is a permutation of 1, 2, ..., N. The last line of the block contains just 0.
The last block consists of just one line containing 0.
Output
The output contains the lines corresponding to the lines with permutations in the input. A line of the output contains Yes if it is possible to marshal the coaches in the order required on the corresponding line of the input. Otherwise it contains No. In addition, there is one empty line after the lines corresponding to one block of the input. There is no line in the output corresponding to the last ``null ' ' block of the input.
Sample Input
5
1 2 3 4 5
5 4 1 2 3
0
6
6 5 4 3 2 1
0
0
Sample Output
Yes
No
Yes
Source
Central Europe 1997
我的解答：
用c++写的
#include <iostream.h>
int function(int a[],int n)
{
int *b=new int[n];
int i,temp,j,k,g;
for(i=0;i <n;i++)
{
temp=a[i];
for(j=i+1,k=0;j <n;j++)
if(a[j] <temp) b[k++]=a[j];
for(g=0;g <k;g++)
if(b[g+1]> b[g]) return 0;
}
return 1;
}
void main()
{
int n,i;
cin> > n;
while(n!=0)
{
int *a=new int[n];
cin> > a[0];
while(a[0]!=0)
{
for(i=1;i <n;i++)
cin> > a[i];
if(function(a,n)) cout < < "Yes " < <endl;
else cout < < "No " < <endl;
cin> > a[0];
}
delete a;
cin> > n;
}
}
[解决办法]
我的代码：
/*
author: TangQiao , Wind @ Beijing Normal University
problem name: Rails
source : PKU Online Judge
problem type: 模拟
problem description: 模拟进栈和出栈过程,看是否能够按题目所说顺利输出.
problem solution: 模拟
date : 2005.9.22
*/
#include <stdio.h>
#include <string.h>
int stack[1010];
int n;
int in[1010], a;
int p_in;
int top;
bool first = true;
int main()
{
int i;
while (1)
{
scanf( "%d ", &n);
if (n == 0) break;
if (first) first = false;
else printf( "\n ");
while (1)
{
scanf( "%d ", in+1);
if (in[1] == 0) break;
for (i=2; i <=n; ++i)
scanf( "%d ", in+i);
memset(stack, 0, sizeof(stack));
p_in = 1;
top = 1;
for (a=1; a <=n; ++a)
{
stack[top++] = a;
while (stack[top-1] == in[p_in] && top > 0 && p_in <=n)
{
top --;
p_in ++;
}
}
if (top ==1) printf( "Yes\n ");
else printf( "No\n ");
}
}
return 0;
}
[解决办法]
直接自己写一个堆栈模拟搜索就可以了吧
#include <iostream>
using namespace std;
int main()
{
int n,in[1000],out[1000],buffer[1000];
while(cin> > n)
{
if(n==0)
break;
while(cin> > out[n-1])
{
if(out[n-1]==0)
break;
for(int i=n-2;i> =0;i--)
cin> > out[i];
for(int j=0;j <n;j++)
{
in[j]=n-j;
buffer[j]=-1;
}
bool canGo=false;
int topIn=n-1,topOut=n-1,topBuffer=-1;
while(1)
{
if(topBuffer==-1&&topIn==-1&&topOut==-1)
{
canGo=true;
break;
}
if(topIn==-1&&(buffer[topBuffer]!=out[topOut])||topBuffer==n-1&&in[topIn]!=out[topOut])
{
canGo=false;
break;
}
if(out[topOut]==in[topIn])
{
topOut--;
topIn--;
}
else if(topBuffer!=-1&&out[topOut]==buffer[topBuffer])
{
topOut--;
topBuffer--;
}
else
{
buffer[++topBuffer]=in[topIn--];
}
}
if(canGo)
cout < < "Yes " < <endl;
else
cout < < "No " < <endl;
}
cout < <endl;
}
return 0;
}
已经ac了的代码
