Uva-12657 Boxes in a Line(双向链表)

来源：互联网发布：飞腾排版软件win7 编辑：程序博客网时间：2024/05/21 17:31
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 100005int 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[n]=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++;    }}
0 0