UVA 12657 Boxes in a Line( 双向链表 )

12657 Boxes in a Line
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

模拟双向链表

逆转的时候记录次数

（逆转之后的操作1相当于2,2相当于1）

注意一下细节就可以过了~

#include<bits/stdc++.h>
using namespace std;
const int N = 100010;
const double PI = acos(-1.0);
const double eps = 1e-6;
typedef long long LL;
typedef pair<double,double> pii;
#define X first
#define Y second

int n , m , L[N] , R[N] , tag ;

void init() {
    tag = 0 ;
    for( int i = 0 ; i <= n ; ++i ){
        L[i] = i-1 , R[i] = i + 1 ;
    }
}
void test() {
    for( int i = 0 ; i <= n ; ++i )
        cout << L[i] <<' ' << R[i] << endl ;
}
void changeNext( int x , int y ) {
    R[ L[x] ] = y ;
    L[ R[y] ] = x ;
    L[y] = L[x] ;
    R[x] = R[y] ;
    L[x] = y ;
    R[y] = x ;
}

void move_l( int x , int y ) {
    if( R[x] == y || x == y ) return ;
    R[ L[x] ] = R[x];
    L[ R[x] ] = L[x];
    L[x] = L[y];
    R[ L[y] ] = x ;
    L[y] = x ;
    R[x] = y ;
}

void move_r( int x , int y ) {
    if( L[x] == y || x == y ) return ;
    R[ L[x] ] = R[x];
    L[ R[x] ] = L[x];
    R[x] = R[y];
    L[ R[y] ] = x ;
    R[y] = x ;
    L[x] = y ;
}
void change( int x , int y ) {
    if( x == y ) return ;
    if( R[x] == y ){
        changeNext(x,y); return ;
    }
    if( R[y] == x ) {
        changeNext(y,x); return ;
    }
    int tmp , l1 , r1 , l2 , r2 ;
    l1 = L[x] , r1 = R[x] , l2 = L[y] , r2 = R[y];
    tmp = L[x] , L[x] = L[y] , L[y] = tmp ;
    tmp = R[x] , R[x] = R[y] , R[y] = tmp ;
    R[l1] = y , L[r1] = y ;
    R[l2] = x , L[r2] = x ;
}

LL cal( int tag ) {
    LL res = 0 ; int cnt = 1 ;
    if( tag && !(n&1) ) tag = 0 ;
    else tag = 1 ;

    for( int i = R[0] ; i <= n ; i = R[i] ){
        if( tag && (cnt&1) ) res += i ;
        if( !tag && !(cnt&1) ) res += i ;
        cnt++;
    }
    return res;
}

void Run() {
    init();
    while(m--){
        int op , x , y ;
        scanf("%d",&op);
        if( op == 1 ) {
            scanf("%d%d",&x,&y);
            if( !tag ) move_l(x,y);
            else move_r(x,y);
        }
        else if( op == 2 ) {
            scanf("%d%d",&x,&y);
            if( !tag ) move_r(x,y);
            else move_l(x,y);
        }
        else if( op == 3 ) {
            scanf("%d%d",&x,&y);
            change(x,y);
        }
        else tag^=1;
//        test();cout << endl ;
    }
    printf("%lld
",cal(tag));
}

int main() {
//    freopen("in.txt","r",stdin);
    int _ , cas = 1 ;
    while( scanf("%d%d",&n,&m) != EOF ) {
        printf("Case %d: ",cas++);
        Run();
    }
}