Example 6-5 Boxes in a line uVa12657

The solution to this problem is a doubly linked list, the data structure itself is not complex, but for four cases of processing is not detailed, mainly reflected in the following points:

1. Classification discussion is not comprehensive, without considering special circumstances (itself does not need to operate, need to interchange two elements adjacent)
2. No consideration of the effect on other operations after the change of State 4
3. No flexible use of mathematical knowledge (courtship only needs to subtract all the odd numbers)

The following post the AC code

#include <cstdio>
#include <algorithm>
const int MAXN = 100000 + 10;
int LEFT[MAXN];
int RIGHT[MAXN];
int S[MAXN];
using namespace Std;
void link (int x,int y) {
Right[x] = y;
Left[y] = x;
}
int main () {
#ifdef DEBUG
Freopen ("6.5.in", "R", stdin);
#endif
int n, m, num = 0;
while (scanf ("%d%d", &n, &m) ==2) {
for (int i = 1; I <= n; i++) {
Right[i]= (i+1);
Left[i]=i-1;
}
Right[0]=1;
Left[0]=n;
int op,x,y;
int INV = 0;
while (m--) {
scanf ("%d", &op);
if (op = = 4) Inv=!inv;
else {
scanf ("%d%d", &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= 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);
}
}
}
}
A long long ans = 0;
int j=0;
for (int i = 1;i<=n; i++) {
J=RIGHT[J];
if (i% 2 = = 1) ans+=j;
}
if (inv && n%2 = = 0) ans = (long Long) n * (n+1)/2-ans;
printf ("Case%d:%lld\n", ++num, ans);
}
return 0;
}

Example 6-5 Boxes in a line uVa12657