Uva-12657 Boxes in a Line(双向链表)

注意操作3是通过数字寻找位置的，不会受是否翻转的影响

#include <cstdio>
#include <algorithm>
using namespace std;

#define BG 100005

int be[BG],ne[BG];
int w=1;
int main()
{
int n,m;
while(scanf("%d%d",&n,&m)!=EOF)
{
int i;
int x,a,b;
int tn,tb;
int inv=0;//use inv to show whether the link has been inverted for odd times.
be[0]=n;
ne[0]=1;
for(i=1;i<n;i++)
ne[i]=i+1;
ne
=0;
for(i=2;i<=n;i++)
be[i]=i-1;
be[1]=0;
for(i=0;i<m;i++)
{
scanf("%d",&x);
if(x==4)
inv=!inv;//fetch the opposite value
else
{
if(x!=3&&inv==1) x=3-x;//if x is 1 or 2,change the direction of the order.
if(x==1)
{
scanf("%d%d",&a,&b);
if(be[b]!=a)
{
ne[be[a]]=ne[a];
be[ne[a]]=be[a];
be[a]=be[b];
ne[a]=b;
ne[be[b]]=a;
be[b]=a;
}
}
else if(x==2)
{
scanf("%d%d",&a,&b);
if(ne[b]!=a)
{
ne[be[a]]=ne[a];
be[ne[a]]=be[a];
ne[a]=ne[b];
be[a]=b;
be[ne[b]]=a;
ne[b]=a;
}
}
else if(x==3)//operation 3 isn't influenced by the inv
{
scanf("%d%d",&a,&b);
ne[be[a]]=b;
be[ne[a]]=b;
tn=ne[b];
tb=be[b];
ne[b]=ne[a];
be[b]=be[a];
ne[tb]=a;
be[tn]=a;
ne[a]=tn;
be[a]=tb;
}

}
// else if(x==4)//waste too much time.
// inv=!inv;
// {
//
// for(i=0;i<=n;i++)
// {
// t=be[i];
// be[i]=ne[i];
// ne[i]=t;
// }
// }
}
i=ne[0];
long long sum=0;
int count=1;
if(n%2==0&&inv==1)
i=ne[ne[0]];
while(i!=0)
{
if(count%2!=0)
sum+=i;
i=ne[i];
count++;
}
printf("Case %d: ",w);
printf("%lld\n",sum);
w++;
}
}