Uva - 12657 - Boxes in a Line

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cctype>
#include <cstring>
#include <string>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <algorithm>
#include <stack>
#include <queue>
#include <bitset>
#include <cassert>

using namespace std;

#define _CRT_SECURE_NO_WARNINGS
const int maxn = 100005;
int n;
int rightP[maxn], leftP[maxn];

void link(int L, int R)
{
rightP[L] = R;
leftP[R] = L;
}

int main()
{
int m, kase = 0;
while (scanf("%d%d", &n, &m) == 2 && kase < 10) {
for (int i = 1; i <= n; i++) { // 初始化n个盒子
leftP[i] = i - 1;
rightP[i] = (i + 1) % (n + 1);
}
rightP[0] = 1;
leftP[0] = n;
int op, x, y, inv = 0;

while (m--) {
scanf("%d", &op);
if (op == 4) { // 4不需要再读取了
inv = !inv;
}
else {
scanf("%d%d", &x, &y);
if (op == 3 && rightP[y] == x) {
swap(x, y);
}
if (op != 3 && inv) {
op = 3 - op;
}
if (op == 1 && x == leftP[y]) {
continue;
}
if (op == 2 && x == rightP[y]) {
continue;
}

int lx = leftP[x], rx = rightP[x], ly = leftP[y], ry = rightP[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 (rightP[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 b = 0;
long long ans = 0;
for (int i = 1; i <= n; i++) { // 从左到右求出奇数和
b = rightP[b];
if (i % 2 == 1) {
ans += b;
}
}
// 如果长度为偶数，并处于反转状态，则用总和减去ans
if (inv && n % 2 == 0) {
ans = (long long)n * (n + 1) / 2 - ans;
}
printf("Case %d: %lld\n", ++kase, ans);
}

return 0;
}