【数组模拟链表（双向）】UVA

Problem Description

给你n和m，分别对应1-n的编号，和m行操作，有四种情况1 X Y：X移动到Y的左边， 2 X Y： X移动到Y的右边， 3 X Y：交换位置， 4 反转整个编号。

代码：

#include<bits/stdc++.h>using namespace std;int lt[100055];//记录i点的左边下标int rt[100055];//记录i点的右边下标void link(int L, int R)//连接L,R,L在左边R在右边的连接方式{    rt[L] = R;    lt[R] = L;}int main(){    int n, m, cas = 1;    while(~scanf("%d %d", &n, &m))    {        for(int i = 1; i <= n; i++)//初始化记录        {            rt[i] = (i + 1) % (n + 1);            lt[i] = i - 1;        }        rt[0] = 1; lt[0] = n;        int op, x, y, inv = 0;        while(m--)        {            scanf("%d", &op);            if(op == 4) inv = !inv;//标记反转            else            {                scanf("%d %d", &x, &y);                if(op != 3 && inv) op = 3 - op; //如果反转 1 变为2, 2 变为1                if(op == 1 && x == lt[y]) continue;                if(op == 2 && x == rt[y]) continue;                int Lx = lt[x], Rx = rt[x], Ly = lt[y], Ry = rt[y];                if(op == 1)//x到y的左边                {                    link(Lx, Rx); link(Ly, x); link(x, y);                }                else if(op == 2)//x到y的右边                {                    link(Lx, Rx); link(x, Ry); link(y, x);                }                else if(op == 3)//交换位置                {                    if(rt[y] == x)//相邻                    {                        link(Ly, x); link(x, y); link(y, Rx);                    }                    else if(rt[x] == y)//相邻                    {                        link(Lx, y); link(y, x); link(x, Ry);                    }                    else//不相邻                    {                        link(Lx, y); link(y, Rx); link(Ly, x); link(x, Ry);                    }                }            }        }        int b = 0;        long long ans = 0;        for(int i = 1; i <= n; i++)        {            b = rt[b];            if(i % 2 != 0) ans += b;        }        if(inv && n % 2 == 0) ans = (long long) n * (n + 1) / 2 - ans;//n为奇数求和结果不变        printf("Case %d: %lld\n", cas++, ans);    }}