UVA - 12657

#include<cstdio>
#include<cstring>
using namespace std;
int left[100005], right[100005];
 
void link(int l, int r)             
{
    left[r] = l;  right[l] = r;
}
 
int main()
{
    int test = 0, op, x, y, temp, n, m;
    while (scanf ("%d %d", &n, &m) != EOF)
    {
        for (int i = 1; i <= n; ++i)
        {
        	right[i] = i + 1;
			left[i] = i - 1;
		} 
        right[n] = 0, left[0] = n, right[0] = 1;
        int reverse = 0;                     
        while (m--)
        {
            scanf ("%d", &op);
            if (op == 4)
			  reverse = !reverse;
            else
            {
                scanf ("%d %d", &x, &y);
                if ( reverse && op != 3)
				{
					op = 3 - op; 				//翻转之后操作一变为把x移到y后面,操作二变为把x移到y前面。
				 } 
                if (right[y] == x && op == 3)   
				{
					temp = x; 
					x = y; 
					y = temp;
				 }     
                    
                if ( (op == 1 && left[y] == x) || (op == 2 && right[y] == x))  continue;
 
                if (op == 1)  
				{
					link (left[x], right[x]);
					link (left[y], x);
					link (x, y);
				 }                   
                    
                else if (op == 2)  
				{
					link (left[x], right[x]);
					 link (x, right[y]);
					  link (y, x);
				}               
                    
                else  if (y == right[x])  
				{
					link (left[x], y);
					link (x, right[y]);
					link (y, x);
				}         
                    
                else                           
                {
                    int ry = right[y], ly = left[y];
                    link (left[x], y);
					 link (y, right[x]);
					  link (ly, x);
					  link (x, ry);
                }
            }
        }
 
        int t = 0; long long ans = 0;
        for (int i = 1; i <= n; ++i)
        {
            t = right[t];
            if (i % 2) ans += t;
        }
        if (n % 2 == 0 && reverse)    //偶数位的和
            ans = (long long) n / 2 * (1 + n) - ans;
        printf ("Case %d: %lld\n", ++test, ans);
    }
    return 0;
}