Uva 12657 （双向链表

最后要求输出奇数位数的和，记录操作4的次数和n的奇偶即可表示出答案
用双向链表来表示关系

#include<cstdio>
#include<iostream>
#include<algorithm>
typedef long long ll;
int righta[100005],lefta[100005];
void link(int L,int R)              //L<--->R
{
    righta[L] = R;lefta[R] = L;
}
using namespace std;
int main()
{
    int m,n,kase = 0;
    while(scanf("%d %d",&n,&m) == 2)
    {
        for(int i = 1;i <= n;i++)
        {
            lefta[i] = i-1;
            righta[i] = (i+1) % (n+1);
        }
        righta[0]=1;lefta[0] = n;//将0->1->2...->n->0这样连起来
        int op,X,Y,inv = 0;//inv来标记链表是否被操作4反转过 根据inv来判断op为1、2时的操作 和输出时的顺序
        while(m--)
        {
            scanf("%d", &op);
            if(op == 4) inv = !inv;
            else
            {
                scanf("%d %d",&X,&Y);
                if(op == 3 && righta[Y] == X) swap(X,Y);
                if(op != 3 && inv) op = 3-op;
                if(op == 1 && X == lefta[Y]) continue;
                if(op == 2 && X == righta[Y]) continue;
                
                int LX = lefta[X],RX = righta[X],LY = lefta[Y],RY = righta[Y];
                if(op == 1) {link(LX,RX);link(LY,X);link(X,Y);}
                else if(op == 2) {link(X,RY);link(LX,RX);link(Y,X);}
                else if(op == 3) 
                {
                    if(righta[X] == Y) {link(LX,Y);link(Y,X);link(X,RY);}
                    else {link(LX,Y);link(Y,RX);link(LY,X);link(X,RY);}
                }
            }
        }
        int b = 0;
        ll ans = 0;
        for(int i= 1;i<=n;i++)
        {
            b=righta[b];
            if(i%2==1) ans+=b;
        }
        if(inv && n%2==0) ans = (ll)n*(n+1)/2-ans;
        printf("Case %d: %lld\n",++kase,ans); 
    }
    return 0;
}