Boxes in a Line UVA - 12657（链表）

You have n boxes in a line on the table numbered 1 . . . n from left to right. Your task is to simulate 4
kinds of commands:
• 1 X Y : move box X to the left to Y (ignore this if X is already the left of Y )
• 2 X Y : move box X to the right to Y (ignore this if X is already the right of Y )
• 3 X Y : swap box X and Y
• 4: reverse the whole line.
Commands are guaranteed to be valid, i.e. X will be not equal to Y .
For example, if n = 6, after executing 1 1 4, the line becomes 2 3 1 4 5 6. Then after executing
2 3 5, the line becomes 2 1 4 5 3 6. Then after executing 3 1 6, the line becomes 2 6 4 5 3 1.
Then after executing 4, then line becomes 1 3 5 4 6 2
Input
There will be at most 10 test cases. Each test case begins with a line containing 2 integers n, m
(1 ≤ n, m ≤ 100, 000). Each of the following m lines contain a command.
Output
For each test case, print the sum of numbers at odd-indexed positions. Positions are numbered 1 to n
from left to right.
Sample Input
6 4
1 1 4
2 3 5
3 1 6
4
6 3
1 1 4
2 3 5
3 1 6
100000 1
4
Sample Output
Case 1: 12
Case 2: 9
Case 3: 2500050000

好麻烦的链表，写swl。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long ll;
int nex[100005],las[100005];

void link(int x,int y)//x插入y左边
{
    nex[x] = y;las[y] = x;
}

int main()
{
    int n,m;
    int kase = 0;
    while(~scanf("%d%d",&n,&m))
    {
        for(int i = 1;i < n;i++)
        {
            nex[i] = i + 1;
            las[i] = i - 1;
        }
        nex[n] = 0;las[n] = n - 1;nex[0] = 1;las[0] = n;
        ll ans = 0;
        int is_reverse = 0;
        for(int i = 1;i <= m;i++)
        {
            int op;scanf("%d",&op);
            if(op == 4)
            {
                is_reverse ^= 1;
                continue;
            }
            if(is_reverse && op <= 2)
            {
                op = 3 - op;
            }
            
            int x,y;scanf("%d%d",&x,&y);
            int lx = las[x],rx = nex[x],ly = las[y],ry = nex[y];
        
            if(op == 1)
            {
                if(ly == x)continue;
                link(lx,rx);link(ly,x);link(x,y);
            }
            else if(op == 2)
            {
                if(ry == x)continue;
                link(lx,rx);link(y,x);link(x,ry);
            }
            else if(op == 3)
            {
                if(rx == y)
                {
                    link(lx,y);link(y,x);link(x,ry);
                }
                else if(lx == y)
                {
                    link(ly,x);link(x,y);link(y,rx);
                }
                else
                {
                    link(ly,x);link(x,ry);link(lx,y);link(y,rx);
                }
            }
        }
        ll sum = 1ll * (1 + n) * n / 2;
        ll ans1 = 0;
        int cnt = 0;
        for(int i = nex[0];i;i = nex[i])
        {
            cnt++;
            if(cnt % 2 == 1)ans1 += i;
        }
        printf("Case %d: ",++kase);
        if(is_reverse && n % 2 == 0)
        {
            printf("%lld\n",sum - ans1);
        }
        else
        {
            printf("%lld\n",ans1);
        }
    }
    return 0;
}

//10 1
//2 1 10

//27 6
//1 7 18
//4
//1 19 4
//1 5 19
//1 15 11
//4