一行箱子----数组模拟双向链表

有如下操作:

1 x y 把x放到y的左边(相邻的左边,如果已经是左边了则忽略)

2 x y 把x放到y的右边(如果已经是右边则忽略)

3 x y 交换x,y的位置

4 反转整条链

 

代码如下：

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#include<cstdio> #include<iostream> using namespace std; #define MAX 100050 int a[MAX],r[MAX],l[MAX]; int n,m; //a的右指针指向b，b的左指针指向a void link(int a,int b) {     r[a] = b;     l[b] = a; } int main() { //    freopen("b.in","r",stdin); //    freopen("out.txt","w",stdout);     int t= 1;     while(scanf("%d%d",&n,&m)!=EOF)     {         r[0]=1; l[n+1]=n;         for(int i=1;i<=n;i++)         {             l[i]=i-1;             r[i]=i+1;         }         int cmd,x,y,xl,xr,yl,yr;         bool flag = false;         for(int i=1;i<=m;i++)         {             scanf("%d",&cmd);             if(cmd==4)                 flag=!flag;             else             {                 if(flag&&cmd!=3)                     cmd = 3-cmd;                 scanf("%d%d",&x,&y);                 if(cmd==1)//把x放在y的左边                 {                     if(r[x]==y&&l[y]==x)                         continue;                     xl = l[x]; xr = r[x];                     yl = l[y];link(xl,xr);                     link(yl,x);link(x,y);                 }                 else if(cmd==2)//把x放在y的右边                 {                     if(l[x]==y&&r[y]==x)                         continue;                     xl = l[x]; xr = r[x];                     yr = r[y]; link(xl,xr);                     link(y,x); link(x,yr);                 }                 else//cmd==3,交换x,y的位置                 {                     //分情况,相邻或不相邻                     //x在y左边                     xl = l[x]; xr = r[x];                     yr = r[y]; yl = l[y];                     if(r[x]==y&&l[y]==x)                         link(xl,y), link(y,x),link(x,yr);                     //x在y右边                     else if(l[x]==y&&r[y]==x)                         link(yl,x),link(x,y),link(y,xr);                     else                     {                         //不相邻                         link(xl,y); link(y,xr);                         link(yl,x); link(x,yr);                     }                 }             }         }         long long ans = 0;         int k = 0;         for(int i=1;i<=n;i++)         {             k = r[k];             if(i&1)                 ans+=k;         }         if(flag&&(n%2==0))//偶数时改变了奇偶顺序         {             long long sum = (long long)(n/2)*(1+n);//此处一定强转不然爆int             ans = sum-ans;         }         printf("Case %d: %lld\n",t++,ans);     }     return 0; }