题目链接：https://vjudge.net/problem/UVA-12657
题目大意
你有n个盒子在桌子上的一条线上从左到右编号为1……n。你的任务是模拟四种操作
- X Y 移动盒子编号X到盒子编号Y的左边（如果X已经在Y的左边了就忽略）
- X Y 移动盒子编号X到盒子编号Y的右边（如果X已经在Y的右边了就忽略）
- X Y 交换盒子编号X与盒子编号Y的位置
- 将整条线反转
操作保证合法，X不等于Y
举一个例子，如果n=6，操作 1 1 4然后就变成了2 3 1 4 5 6；再操作 2 3 5就变成了 2 1 4 5 3 6；再操作 3 1 6 就变成 2 6 4 5 3 1；最后操作4，就变成了 1 3 5 4 6 2
输入
最多有10组数据，每个数据会包含两个整数n,m（1≤n,m<100,000）, 接下来是m行数据，表示操作。
输出
对于每组数据，输出他们奇数位置的编号的和。
翻译搬运自洛谷。
分析
静态链表模板题。代码如下
1 #include <bits/stdc++.h>
2 using namespace std;
3
4 #define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
5 #define Rep(i,n) for (int i = 0; i < (n); ++i)
6 #define For(i,s,t) for (int i = (s); i <= (t); ++i)
7 #define rFor(i,t,s) for (int i = (t); i >= (s); --i)
8 #define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
9 #define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
10 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
11 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)
12
13 #define pr(x) cout << #x << " = " << x << " "
14 #define prln(x) cout << #x << " = " << x << endl
15
16 #define LOWBIT(x) ((x)&(-x))
17
18 #define ALL(x) x.begin(),x.end()
19 #define INS(x) inserter(x,x.begin())
20
21 #define ms0(a) memset(a,0,sizeof(a))
22 #define msI(a) memset(a,inf,sizeof(a))
23 #define msM(a) memset(a,-1,sizeof(a))
24
25 #define MP make_pair
26 #define PB push_back
27 #define ft first
28 #define sd second
29
30 template<typename T1, typename T2>
31 istream &operator>>(istream &in, pair<T1, T2> &p) {
32 in >> p.first >> p.second;
33 return in;
34 }
35
36 template<typename T>
37 istream &operator>>(istream &in, vector<T> &v) {
38 for (auto &x: v)
39 in >> x;
40 return in;
41 }
42
43 template<typename T1, typename T2>
44 ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
45 out << "[" << p.first << ", " << p.second << "]" << "\n";
46 return out;
47 }
48
49 inline int gc(){
50 static const int BUF = 1e7;
51 static char buf[BUF], *bg = buf + BUF, *ed = bg;
52
53 if(bg == ed) fread(bg = buf, 1, BUF, stdin);
54 return *bg++;
55 }
56
57 inline int ri(){
58 int x = 0, f = 1, c = gc();
59 for(; c<48||c>57; f = c=='-'?-1:f, c=gc());
60 for(; c>47&&c<58; x = x*10 + c - 48, c=gc());
61 return x*f;
62 }
63
64 typedef long long LL;
65 typedef unsigned long long uLL;
66 typedef pair< double, double > PDD;
67 typedef pair< int, int > PII;
68 typedef pair< string, int > PSI;
69 typedef set< int > SI;
70 typedef vector< int > VI;
71 typedef vector< VI > VVI;
72 typedef vector< PII > VPII;
73 typedef map< int, int > MII;
74 typedef multimap< int, int > MMII;
75 typedef unordered_map< int, int > uMII;
76 typedef pair< LL, LL > PLL;
77 typedef vector< LL > VL;
78 typedef vector< VL > VVL;
79 typedef priority_queue< int > PQIMax;
80 typedef priority_queue< int, VI, greater< int > > PQIMin;
81 const double EPS = 1e-10;
82 const LL inf = 0x7fffffff;
83 const LL infLL = 0x7fffffffffffffffLL;
84 const LL mod = 1e9 + 7;
85 const int maxN = 1e5 + 7;
86 const LL ONE = 1;
87 const LL evenBits = 0xaaaaaaaaaaaaaaaa;
88 const LL oddBits = 0x5555555555555555;
89
90 struct StaticList{
91 int sl[maxN], nt[maxN], pv[maxN];
92 int *next = nt, *prev = pv;
93 int sz;
94
95 void init(int x) {
96 sz = x;
97 For(i, 0, sz) {
98 if(i) sl[i] = i;
99 prev[i] = (sz + i) % (sz + 1);
100 next[i] = (i + 1) % (sz + 1);
101 }
102 }
103
104 // 把 x 插到 y 前面
105 void move(int x, int y) {
106 if(y == x || next[x] == y) return;
107 next[prev[x]] = next[x];
108 prev[next[x]] = prev[x];
109
110 next[prev[y]] = x;
111 prev[x] = prev[y];
112 next[x] = y;
113 prev[y] = x;
114 }
115
116 void Swap(int x, int y) {
117 // 如果两个节点相邻，要单独讨论
118 if(next[x] == y || prev[x] == y) {
119 if(prev[x] == y) swap(x, y);
120 swap(next[prev[x]], prev[next[y]]);
121 next[x] = next[y];
122 prev[y] = prev[x];
123 prev[x] = y;
124 next[y] = x;
125 return;
126 }
127
128 swap(next[prev[x]], next[prev[y]]);
129 swap(prev[next[x]], prev[next[y]]);
130 swap(prev[x], prev[y]);
131 swap(next[x], next[y]);
132 }
133
134 void reverse(){
135 swap(prev, next);
136 }
137
138 LL getAns() {
139 LL ret = 0;
140 int p = next[0];
141
142 while(p != 0) {
143 ret += sl[p];
144 p = next[p];
145 if(p == 0) break;
146 p = next[p];
147 }
148
149 return ret;
150 }
151
152 void print() {
153 cout << "SL: ";
154 For(i, 0, sz) cout << " " << sl[i];
155 cout << endl;
156
157 cout << "prev:";
158 For(i, 0, sz) cout << " " << prev[i];
159 cout << endl;
160
161 cout << "next:";
162 For(i, 0, sz) cout << " " << next[i];
163 cout << endl;
164
165 }
166 }SL;
167
168 int n, m, T;
169
170 int main(){
171 //freopen("MyOutput.txt","w",stdout);
172 INIT();
173 while(cin >> n >> m) {
174 ++T;
175 SL.init(n);
176 Rep(i, m) {
177 int t, x, y;
178 cin >> t;
179 if(t == 4) SL.reverse();
180 else {
181 cin >> x >> y;
182 switch(t) {
183 case 1:{
184 SL.move(x, y);
185 break;
186 }
187 case 2:{
188 SL.move(x, SL.next[y]);
189 break;
190 }
191 case 3:{
192 SL.Swap(x, y);
193 break;
194 }
195 }
196 }
197 //SL.print();
198 }
199 //SL.print();
200 cout << "Case " << T << ": " << SL.getAns() << endl;
201 }
202 return 0;
203 }
View Code