UVa 12657 Boxes in a Line 双向链表

题目大意：

四个操作

1 X Y 把X移到Y的左边

2 X Y  把X移到Y的右边

3 X Y 交换

4 翻转

用双向链表

前两个没什么好讲的，

第三个操作要注意相邻元素之间的交换

最后一个可以用标记，然后最后计算的时候判断头尾即可（翻转后1，2 操作交换）

具体代码实现如下：

   
     

    
      
#include<cstdio>
    
      
using namespace std;
    
      

    
      
int left[100010], right[100010];
    
      

    
      
int main() {
    
      
 int n, m, times(0);
    
      
 while (scanf("%d%d", &n, &m) == 2) {
    
      
 for (int i = 0; i != n + 3; ++i) { //初始化左右
    
      
 left[i] = i - 1;
    
      
 right[i] = i + 1;
    
      
 }
    
      
 int op, x, y, chan(0);
    
      
 while (m--) {
    
      
 scanf("%d", &op);
    
      
 if (op == 4) chan = !chan;
    
      
 else {
    
      
 scanf("%d%d", &x, &y);
    
      
 if (chan && (op == 1 || op == 2)) {
    
      
 if (op == 1) op = 2;
    
      
 else op = 1;
    
      
 }
    
      
 if (op == 1) {
    
      
 left[right[x]] = left[x];
    
      
 right[left[x]] = right[x];
    
      
 left[x] = left[y];
    
      
 right[x] = y;
    
      
 right[left[y]] = x;
    
      
 left[y] = x;
    
      
 }else if (op == 2) {
    
      
 left[right[x]] = left[x];
    
      
 right[left[x]] = right[x];
    
      
 right[x] = right[y];
    
      
 left[x] = y;
    
      
 left[right[y]] = x;
    
      
 right[y] = x;
    
      
 }else {
    
      
 int lx = left[x], ly = left[y], 
    
      
 rx = right[x], ry = right[y];
    
      
 if (rx == y) {
    
      
 left[y] = lx;
    
      
 right[x] = ry;
    
      
 right[y] = left[ry] = x;
    
      
 left[x] = right[lx] = y;
    
      
 } else if (lx == y) {
    
      
 left[x] = ly;
    
      
 right[y] = rx;
    
      
 right[x] = left[rx] = y;
    
      
 left[y] = right[ly] = x;
    
      
 } else {
    
      
 right[lx] = left[rx] = y;
    
      
 right[ly] = left[ry] = x;
    
      
 left[x] = ly;
    
      
 right[x] = ry;
    
      
 left[y] = lx;
    
      
 right[y] = rx;
    
      
 }
    
      
 }
    
      
 }
    
      
 }
    
      
 long long sum(0);
    
      
 int b(0);
    
      
 if (chan) {
    
      
 b = n + 1;
    
      
 for (int i = 1; i != n + 1; ++i) {
    
      
 b = left[b];
    
      
 if (i % 2) sum += b;
    
      
 // printf("%d ", b); //*
    
      
 }
    
      
 } else {
    
      
 for (int i = 1; i != n + 1; ++i) {
    
      
 b = right[b];
    
      
 if (i % 2) sum += b;
    
      
 // printf("%d ", b); //*
    
      
 }
    
      
 }
    
      
 printf("Case %d: %lld\n", ++times, sum);
    
      
 }
    
      
 return 0; 
    
      
} 
   
     

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