移动盒子，紫书P144UVa12657

本题是双向链表的典型应用。用数组模拟链表在算法竞赛中是非常实用并且高效的做法。本题刘代码中最值得学习的地方是link函数的使用，缩短了代码的复杂度。至于本题的思路比较直接，照着题目的要求操作即可。

// UVa12657 Boxes in a Line// Rujia Liu#include<cstdio>#include<algorithm>using namespace std;const int maxn = 100000 + 5;int n, left[maxn], right[maxn];inline void link(int L, int R) {  right[L] = R; left[R] = L;}int main() {  int m, kase = 0;  while(scanf("%d%d", &n, &m) == 2) {    for(int i = 1; i <= n; i++) {      left[i] = i-1;      right[i] = (i+1) % (n+1);    }    right[0] = 1; left[0] = n;    int op, X, Y, 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); }        }      }    }    int b = 0;    long long ans = 0;    for(int i = 1; i <= n; i++) {      b = right[b];      if(i % 2 == 1) ans += b;    }    if(inv && n % 2 == 0) ans = (long long)n*(n+1)/2 - ans;    printf("Case %d: %lld\n", ++kase, ans);  }  return 0;}