UVA12657 Boxes in a line "bidirectional linked list" "Array emulation"

You are n boxes in a line on the table numbered 1 ... Your task is to simulate 4

Kinds of commands:

? 1 x Y:move box X to the left and Y (ignore this if X are already the left of Y)
? 2 x Y:move box X to the right and Y (ignore this if X are already the right of Y)
? 3 x Y:swap box X and Y

? 4:reverse the whole line.

Commands is guaranteed to be valid, i.e. X would be is 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'll is at the most 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 is 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 3:2,500,050,000

You have a row of boxes, numbered from left to right 1~n, now there are 4 kinds of operations.

1 x y means move the X box to the left of the Y box.

2 x y means move the X box to the right of the Y box.

3 x y means the position of the X box and the Y box are exchanged

4 Flip all the boxes in order.

The last question is the number of boxes in odd position after m operation.

Idea: The best way to use a doubly linked list. This is simulated using an array method, using Left[i] and Right[i]

The box number for the left and right of the box numbered I, respectively (0 means no box). by simulating

The linked list method changes the connection order.

#include <iostream> #include <algorithm> #include <cstdio> #include <cstring>using namespace    Std;int left[100010],right[100010];void Link (int l,int R) {right[l] = R; Left[r] = L;}    int main () {int n,m,kase = 0;            while (CIN >> n >> m) {for (int i = 1; I <= n; i++) {left[i] = i-1;        Right[i] = (i+1)% (n+1);        } right[0] = 1;        Left[0] = n;        int INV = 0,x,y;            while (m--) {int op;            Cin >> op;            if (op = = 4) INV =!INV;                else {cin >> 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,rx,ly,RY;                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);        }}}} int num = 0;        A long long ans = 0;            for (int i = 1; I <= n; i++) {num = Right[num];if (i&1) ans + = num; } if (INV &&!) (        n&1) ans = (long Long) n (n+1)/2-ans;    cout << "Case" << ++kase << ":" << ans << endl; } return 0;}

UVA12657 Boxes in a line "bidirectional linked list" "Array emulation"