Description
Your friend, Jackson is invited to a TV show called SuperMemo in which the participant is told to play a memorizing game. At first, the host tells the participant a sequence of numbers, {A1, A2, ... An}. Then the host performs a series of operations and queries on the sequence which consists:
- ADD x y D: Add D to each number in sub-sequence {Ax ... Ay}. For example, performing "ADD 2 4 1" on {1, 2, 3, 4, 5} results in {1, 3, 4, 5, 5}
- REVERSE x y: reverse the sub-sequence {Ax ... Ay}. For example, performing "REVERSE 2 4" on {1, 2, 3, 4, 5} results in {1, 4, 3, 2, 5}
- REVOLVE x y T: rotate sub-sequence {Ax ... Ay} T times. For example, performing "REVOLVE 2 4 2" on {1, 2, 3, 4, 5} results in {1, 3, 4, 2, 5}
- INSERT x P: insert P after Ax. For example, performing "INSERT 2 4" on {1, 2, 3, 4, 5} results in {1, 2, 4, 3, 4, 5}
- DELETE x: delete Ax. For example, performing "DELETE 2" on {1, 2, 3, 4, 5} results in {1, 3, 4, 5}
- MIN x y: query the participant what is the minimum number in sub-sequence {Ax ... Ay}. For example, the correct answer to "MIN 2 4" on {1, 2, 3, 4, 5} is 2
To make the show more interesting, the participant is granted a chance to turn to someone else that means when Jackson feels difficult in answering a query he may call you for help. You task is to watch the TV show and write a program giving the correct answer to each query in order to assist Jackson whenever he calls.
Input
The first line contains n (n ≤ 100000).
The following n lines describe the sequence.
Then follows M (M ≤ 100000), the numbers of operations and queries.
The following M lines describe the operations and queries.
Output
For each "MIN" query, output the correct answer.
Sample Input
5 1 2 3 4 5 2 ADD 2 4 1 MIN 4 5
Sample Output
5
Source
POJ Founder Monthly Contest – 2008.04.13, Yao Jinyu
【分析】
真实蛋疼了。。
第一次写的时候居然犯了一个傻逼到极点的错误。。。。
翻转部分不会，是看别人的。不过感觉应该还有不同的写法。
代码写得比较丑，主要是尝试了引用。
1 #include <iostream>
2 #include <cstdio>
3 #include <algorithm>
4 #include <cstring>
5 #include <vector>
6 #include <utility>
7 #include <iomanip>
8 #include <string>
9 #include <cmath>
10 #include <queue>
11 #include <assert.h>
12 #include <map>
13 #include <ctime>
14 #include <cstdlib>
15 #define LOCAL
16 const int MAXN = 200000 + 10;
17 const int INF = 100000000;
18 const int SIZE = 450;
19 const int maxnode = 0x7fffffff + 10;
20 using namespace std;
21 typedef struct sp_node * Node;
22 struct SPLAY{
23 struct Node{
24 int val, Min;//分别为值,最小值,大小和lazy下标
25 int size, lazy;
26 bool turn;
27 Node *parent, *ch[2];
28
29 //初始化
30 void NEW(int x){
31 Min = val = x;
32 size = 1;
33 lazy = turn = 0;
34 parent = ch[0] = ch[1] = NULL;
35 }
36 int cmp(){
37 if (parent->ch[0] == this) return 0;
38 if (parent->ch[1] == this) return 1;
39 }
40 //还是不要写在里面了..
41 /*void pushdown(){
42
43 }
44 void update(){
45 size = 1;
46 if (x->ch[0] != NULL) size += x->ch[0]->size;
47 if (x->ch[1] != NULL) size += x->ch[0]->size;
48 }
49 */
50 }mem[MAXN], *root;//mem为静态数组
51 int tot;
52
53 int get(Node *&x){return (x == NULL ? 0 : x->size);}
54 //————————————————————这一部分不能卸载里面
55 //更新
56 void update(Node *&x){
57 if (x == NULL) return;
58 x->size = 1;
59 x->Min = x->val;
60 if (x->ch[0] != NULL) {x->Min = min(x->Min, x->ch[0]->Min);x->size += x->ch[0]->size;}
61 if (x->ch[1] != NULL) {x->Min = min(x->Min, x->ch[1]->Min);x->size += x->ch[1]->size;}
62 }
63 void pushdown(Node *&x){//标记下传
64 if (x == NULL) return;
65 if (x->lazy){
66 int tmp = x->lazy;
67 x->val += tmp;
68 if (x->ch[0] != NULL) {x->ch[0]->lazy += tmp;x->ch[0]->Min += tmp;}
69 if (x->ch[1] != NULL) {x->ch[1]->lazy += tmp;x->ch[1]->Min += tmp;}
70 x->lazy = 0;
71 }
72 if (x->turn){//翻转
73 swap(x->ch[0], x->ch[1]);//swap内部用什么实现的呢？
74
75 if (x->ch[0] != NULL) x->ch[0]->turn ^= 1;
76 if (x->ch[1] != NULL) x->ch[1]->turn ^= 1;
77 x->turn = 0;
78 }
79 }
80
81 //————————————————————————————
82 //旋转，1为右旋, 0为左旋
83 void Rotate(Node *&x, int d){//不用引用也可以
84 Node *y = x->parent;
85 pushdown(y);//注意顺序
86 pushdown(x);
87 pushdown(x->ch[d]);//这个不要忘了
88
89 y->ch[d ^ 1] = x->ch[d];
90 if (x->ch[d] != NULL) x->ch[d]->parent = y;
91 x->parent = y->parent;
92 if (y->parent != NULL){
93 if (y->parent->ch[0] == y) y->parent->ch[0] = x;
94 else y->parent->ch[1] = x;
95 }
96 x->ch[d] = y;
97 y->parent = x;
98 update(y);
99 if (y == root) root = x;//如果是引用要小心root被传下去
100 }
101 void debug(){
102 Node *p = new Node;
103 root = new Node;
104 root->ch[1] = p;
105 p->parent = root;
106 printf("%d", p->cmp());
107 }
108 //注意我写的这个跟下午那个不一样，下午那个好麻烦
109 void splay(Node *&x, Node *&y){
110 pushdown(x);
111 while (x != y){
112 if(x->parent == y){
113 if (y->ch[0] == x) Rotate(x, 1);
114 else Rotate(x,0);
115 break;
116 }else{
117 Node *y = x->parent, *z = y->parent;//注意一定要这样弄一下
118 if (z->ch[0] == y)
119 if (y->ch[0] == x) Rotate(y,1),Rotate(x, 1);
120 else Rotate(x, 0), Rotate(x, 1);
121 else if (y->ch[1] == x) Rotate(y, 0), Rotate(x, 0);
122 else Rotate(x, 1), Rotate(x, 0);
123 if (z == y) break;
124 }
125 update(x);
126 }
127 update(x);
128 }
129 //寻找第k大
130 void find(Node *&t, int k){
131 int tmp;
132 Node *p = root;
133 while (1){
134 pushdown(p);
135 tmp = get(p->ch[0]);
136 if (k == (tmp + 1)) break;
137 if (k <= tmp) p = p->ch[0];
138 else {k -= tmp + 1, p = p->ch[1];}
139 }
140 pushdown(p);
141 splay(p, t);
142 }
143 //插入操作
144 void Insert(int pos, int val){
145 //还是卡出来
146 find(root, pos + 1);
147 find(root->ch[1], pos + 2);
148 Node *t = &mem[tot++], *x = root->ch[1];
149 pushdown(root);
150 pushdown(x);
151 t->NEW(val);
152 //直接拆开放中间,放在root->ch[1]->ch[0]应该也可以
153 t->ch[1] = x;
154 x->parent = t;
155 root->ch[1] = t;
156 t->parent = root;
157 splay(x, root);
158 }
159 //区间加
160 void Add(int l, int r, int x){
161 find(root, l);
162 find(root->ch[1] , r + 2);
163 Node *t = root->ch[1]->ch[0];
164 pushdown(t);
165 update(t);
166 t->Min += x;
167 t->lazy += x;
168 splay(t, root);
169 }
170 //翻转操作
171 void Reverse(int l, int r){
172 find(root, l);
173 find(root->ch[1], r + 2);
174 root->ch[1]->ch[0]->turn ^= 1;
175 Node *x = root->ch[1]->ch[0];
176 splay(x, root);
177 }
178 //交换操作,这一段我是看别人的....
179 void Revolve(int l, int r, int t){
180 Node *p1, *p2;
181
182 find(root, l);
183 find(root->ch[1], r + 2);
184 find(root->ch[1]->ch[0], r + 1 - t);
185
186 p1 = root->ch[1]->ch[0];
187 pushdown(p1);
188 p2 = p1->ch[1];
189 p1->ch[1] = NULL;
190 find(root->ch[1]->ch[0], l + 1);
191 p1 = root->ch[1]->ch[0];
192 pushdown(p1);
193 p1->ch[0] = p2;
194 p2->parent = p1;
195 splay(p2, root);
196 }
197 int getMin(int l, int r){
198 find(root, l);
199 find(root->ch[1], r + 2);
200 Node *x = root->ch[1];
201 pushdown(x);
202 x = x->ch[0];
203 pushdown(x);
204 update(x);
205 return x->Min;
206 }
207 void Erase(int pos){//删除
208 find(root, pos);
209 find(root->ch[1], pos + 2);
210 pushdown(root->ch[1]);
211 root->ch[1]->ch[0] = NULL;
212 Node *x = root->ch[1];
213 splay(x, root);
214 }
215 void init(){
216 //注意还要加一个正无穷
217 tot = 0;
218 root = &mem[tot++];
219 root->NEW(INF);
220 root->ch[1] = &mem[tot++];
221 root->ch[1]->NEW(INF);
222 }
223 void print(Node *t){
224 if (t == NULL) return;
225 print(t->ch[0]);
226 printf("%d ", t->val);
227 if (t->parent != NULL) printf("%d\n", t->parent->val);
228 print(t->ch[1]);
229 }
230 }A;
231
232 int n, m;
233 //注意在这里为了防止掉到0，所以每个位置都要+1
234 void init(){//初始化
235 A.init();
236 scanf("%d", &n);
237 for (int i = 0 ; i < n; i++){
238 int x;
239 scanf("%d", &x);
240 A.Insert(i, x);
241 //printf("%d\n", A.root->val);
242 }
243 //A.print(A.root);
244 }
245 void work(){
246 scanf("%d", &m);
247 for (int i = 1; i <= m; i++){//询问,按顺序来。。
248 char str[20];
249 scanf("%s", str);
250 if (str[0] == ‘A‘){
251 int l, r, x;
252 scanf("%d%d%d", &l, &r, &x);
253 A.Add(l, r, x);
254 }else if (str[0] == ‘R‘){
255 int l, r;
256 scanf("%d%d", &l, &r);
257 if (str[3] == ‘E‘) A.Reverse(l, r);
258 else{
259 int x, len;
260 scanf("%d", &x);
261 len = r - l + 1;
262 x = (x % len + len) % len;
263 if (x == 0 || l == r) continue;//我开始居然傻逼的写在前面....
264 A.Revolve(l, r, x);
265 }
266 }else if (str[0] == ‘I‘){
267 int l, x;
268 scanf("%d%d", &l, &x);
269 A.Insert(l, x);
270 }else if (str[0] == ‘D‘){
271 int l;
272 scanf("%d", &l);
273 A.Erase(l);
274 }else if (str[0] == ‘M‘){
275 int l, r;
276 scanf("%d%d", &l, &r);
277 printf("%d\n", A.getMin(l, r));
278 }
279 }
280 }
281 void debug(){
282
283 }
284
285 int main(){
286
287 init();
288 work();
289 //debug();
290 //A.debug();
291 return 0;
292 }
时间： 2024-11-05 11:51:54