例题6-5 Boxes in a line uVa12657

这道题目的解决方案是双向链表，数据结构本身并不复杂，但对于四种情况的处理不够细致，主要体现在以下几点：

1. 分类讨论不全面，没有考虑特殊情况（本身不需要操作，需要互换的两元素相邻）
2. 没有考虑状态4改变后对其他操作的影响
3. 没有灵活运用数学知识（求偶只需要全部减去奇数即可）

以下贴出AC代码

#include <cstdio>
#include <algorithm>
const int maxn = 100000 + 10;
int left[maxn];
int right[maxn];
int s[maxn];
using namespace std;
void link(int x,int y){
    right[x] = y;
    left[y] = x;
}
int main(){
    #ifdef DEBUG
    freopen("6.5.in","r",stdin);
    #endif
    int n, m ,num = 0;
    while(scanf("%d %d", &n, &m)==2){
        for(int i = 1; i <= n; i++){
            right[i]= (i+1);
            left[i]=i-1;
        }
        right[0]=1;
        left[0]=n;
        int op,X,Y;
        int inv = 0;
        while(m--){
            scanf("%d",&op);
            if(op == 4) inv=!inv;
            else {
                scanf("%d%d",&X, &Y);
                if(op == 3 && right[Y] == X) swap(X,Y);
                if(op != 3 && inv) op = 3 - op;
                if(op == 1 && X == left[Y]) continue;
                if(op == 2 && X == right[Y]) continue;

                int LX= left[X],RX = right[X],LY = left[Y], RY = right[Y];
                if(op == 1){
                    link(LX,RX);link(LY,X);link(X,Y);
                }
                else if(op == 2){
                    link(LX,RX);link(Y,X);link(X,RY);
                }
                else if(op == 3){
                    if(right[X] == Y){
                        link(LX, Y);link(Y, X); link(X, RY);
                    }
                    else{
                         link(LX, Y);link(Y, RX); link(LY, X);link(X,RY);                      
                    }
                }
            }
        }
        long long ans = 0;
        int j=0;
        for(int i = 1;i<=n; i++){
            j=right[j];
            if(i % 2 == 1)   ans+=j;
        }
        if(inv && n %2 == 0) ans =(long long )n * (n+1) /2 -ans;
        printf("Case %d: %lld ", ++num, ans);
    }
    return 0;
}