SuperMemo
| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 12788 | Accepted: 3986 | |
| Case Time Limit: 2000MS | ||
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
这也许是我目前见过最难的平衡树的题目了，相信把这题搞透之后，平衡树也就差不多了 tips：这题我交了50次才对 ~~~
思路其实很简单，主要是程序实现和各种细节。
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <cmath>
5 #include <algorithm>
6 using namespace std;
7 const int maxn=200010;
8 const int INF=2147483647;
9 int Min[maxn],key[maxn],ch[maxn][2],fa[maxn],flip[maxn],Mark[maxn],size[maxn],root,cnt;
10
11 void Push_up(int p)
12 {
13 if(!p)return;
14 Min[p]=min(key[p],min(Min[ch[p][0]],Min[ch[p][1]]));
15 size[p]=size[ch[p][0]]+size[ch[p][1]]+1;
16 }
17 void Add(int p,int d)
18 {
19 if(!p)return;
20 Min[p]+=d;
21 key[p]+=d;
22 Mark[p]+=d;
23 }
24 void Flip(int p)
25 {
26 if(!p)return;
27 swap(ch[p][0],ch[p][1]);
28 flip[p]^=1;
29 }
30 void Push_down(int p)
31 {
32 if(Mark[p]){
33 Add(ch[p][0],Mark[p]);
34 Add(ch[p][1],Mark[p]);
35 Mark[p]=0;
36 }
37 if(flip[p]){
38 Flip(ch[p][0]);
39 Flip(ch[p][1]);
40 flip[p]=0;
41 }
42 }
43 void Rotate(int x)
44 {
45 int y=fa[x],g=fa[y],c=ch[y][1]==x;
46 ch[y][c]=ch[x][c^1];ch[x][c^1]=y;
47 fa[ch[y][c]]=y;fa[y]=x;fa[x]=g;
48 Push_up(y);Push_up(x);
49 if(g)
50 ch[g][ch[g][1]==y]=x;
51 }
52
53 void Splay(int x,int g=0)
54 {
55 for(int y;(y=fa[x])!=g;Rotate(x))
56 if(fa[y]!=g)
57 Rotate((ch[fa[y]][1]==y)==(ch[y][1]==x)?y:x);
58 if(g==0)
59 root=x;
60 }
61
62 void Bring_up(int x,int g=0)
63 {
64 int p=root;
65 while(true){
66 Push_down(p);
67 if(size[ch[p][0]]+1==x)break;
68 if(size[ch[p][0]]>=x)
69 p=ch[p][0];
70 else{
71 x-=size[ch[p][0]]+1;
72 p=ch[p][1];
73 }
74 }
75 Splay(p,g);
76 }
77 #define Go ch[ch[root][1]][0]
78 int main()
79 {
80 int n,p,l,r,x,Q;
81 scanf("%d",&n);
82 Min[0]=INF;size[0]=0;
83 for(int i=0;i<=n+1;i++)
84 {
85 if(i!=0&&i!=n+1)
86 scanf("%d",&x);
87 else
88 x=INF;
89 p=++cnt;
90 key[p]=Min[p]=x;
91 size[p]=1;
92 if(root)
93 ch[root][1]=p,fa[p]=root,Rotate(p);
94 root=p;
95 }
96 char s[15];
97 scanf("%d",&Q);
98 while(Q--)
99 {
100 scanf("%s",s);
101 if(!strcmp(s,"MIN")){
102 scanf("%d%d",&l,&r);l++;r++;
103 Bring_up(l-1);
104 Bring_up(r+1,root);
105 Push_down(root);
106 Push_down(ch[root][1]);
107 printf("%d\n",Min[Go]);
108 }
109 else if(!strcmp(s,"ADD")){
110 scanf("%d%d%d",&l,&r,&x);
111 l++;r++;
112 Bring_up(l-1);
113 Bring_up(r+1,root);
114 Add(Go,x);
115 Push_up(ch[root][1]);
116 Push_up(root);
117 }
118 else if(!strcmp(s,"INSERT")){
119 scanf("%d%d",&l,&x);l++;
120 Bring_up(l);
121 Bring_up(l+1,root);
122 Go=++cnt;
123 key[Go]=Min[Go]=x;
124 size[Go]=1;
125 if(root)
126 fa[Go]=ch[root][1];
127 else
128 root=Go;
129 Push_up(ch[root][1]);
130 Push_up(root);
131 }
132 else if(!strcmp(s,"DELETE")){
133 scanf("%d",&l);l++;
134 Bring_up(l-1);
135 Bring_up(l+1,root);
136 Go=0;
137 Push_up(ch[root][1]);
138 Push_up(root);
139 }
140 else if(!strcmp(s,"REVERSE")){
141 scanf("%d%d",&l,&r);l++;r++;
142 Bring_up(l-1);
143 Bring_up(r+1,root);
144 Flip(Go);
145 }
146 else if(!strcmp(s,"REVOLVE")){
147 scanf("%d%d%d",&l,&r,&x);l++;r++;
148 x%=r-l+1;
149 if(x==0)continue;
150 Bring_up(r-x);
151 Bring_up(r+1,root);
152 int tmp=Go;Go=0;
153 Push_up(ch[root][1]);
154 Push_up(root);
155 Bring_up(l-1);
156 Bring_up(l,root);
157 Go=tmp;fa[Go]=ch[root][1];
158 Push_up(ch[root][1]);
159 Push_up(root);
160 }
161 }
162 return 0;
163 }
时间： 2024-12-16 05:46:03