UVa 12657 Boxes in a line (application doubly linked list)

Boxes in a line

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

InputThere'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 Input6 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 OutputCase 1:12
Case 2:9
Case 3:2500050000

Test instructions start with n boxes in order of 1 to N to perform m operations on each of these boxes for one of the x, Y reverse order of the two to the left of Y.

After the operation is complete, all odd digits of the original box sequence number and;

Direct simulation will definitely time out with the list in the STL also time-out can only use the array itself to simulate a doubly linked list le[i],ri[i] respectively, the box to the left of the first box and the right box in the ordinal code has a stare

#include <cstdio> #include <cstring>using namespace std;const int n = 100005;int le[n], ri[n], N, M;typedef long Long ll;void link (int l, int r)//connect L and R.  L on the left {le[r] = l; Ri[l] = r;}    int main () {int cas = 0, op, x, Y, t; while (scanf ("%d%d", &n, &m)! = EOF) {for (int i = 1; I <= n; ++i) Ri[i] = i + 1, le[i]        = I-1;        Ri[n] = 0, le[0] = n, ri[0] = 1;                      int flag = 0;            Infers whether to flip while (m--) {scanf ("%d", &op);            if (op = = 4) flag =!flag;                else {scanf ("%d%d", &x, &y); if (flag && OP! = 3) op = 3-op;  After flipping the move operation is reversed if (ri[y] = = x && op = = 3)//Convenient to later infer whether the interchange is adjacent t = x, x = y, y =                T if (op = = 1 && le[y] = = x) | |  (OP = = 2 && Ri[y] = = x))                Continue           if (op = = 1)//x move to the right of Y         Link (le[x], ri[x]), Link (le[y], x), link (x, y);                else if (op = = 2)//x moves to the left of Y Link (le[x], ri[x]), link (x, ri[y]), link (y, x); else if (y = = ri[x])//op==3&&x,y adjacent link (le[x], y), link (x, ri[y]), link                (y, x);                    else//nonadjacent {int ry = Ri[y], ly = le[y];                Link (le[x], y), Link (y, ri[x]), Link (ly, x), link (x, ry); }}} t = 0;        ll ans = 0;            for (int i = 1; I <= n; ++i) {t = ri[t];        if (i% 2) ans + = t;        if (n 2 = = 0 && flag)//n is an even number and flipped over so that it is just even digits and ans = (ll) N/2 * (1 + N)-ans;    printf ("Case%d:%lld\n", ++cas, ans); } return 0;}

Copyright notice: This article Bo Master original article. Blog, not reproduced without consent.

UVa 12657 Boxes in a line (application doubly linked list)