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：

#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
",t++,ans);
    }
    return 0;
}