Poj 3580-SuperMemo Splay

题目：http://poj.org/problem?id=3580

 

SuperMemo

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 13105		Accepted: 4104
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, ... A_n}. Then the host performs a series of operations and queries on the sequence which consists:

1. ADD x y D: Add D to each number in sub-sequence {A_x ... A_y}. For example, performing "ADD 2 4 1" on {1, 2, 3, 4, 5} results in {1, 3, 4, 5, 5}
2. REVERSE x y: reverse the sub-sequence {A_x ... A_y}. For example, performing "REVERSE 2 4" on {1, 2, 3, 4, 5} results in {1, 4, 3, 2, 5}
3. REVOLVE x y T: rotate sub-sequence {A_x ... A_y} T times. For example, performing "REVOLVE 2 4 2" on {1, 2, 3, 4, 5} results in {1, 3, 4, 2, 5}
4. INSERT x P: insert P after A_x. For example, performing "INSERT 2 4" on {1, 2, 3, 4, 5} results in {1, 2, 4, 3, 4, 5}
5. DELETE x: delete A_x. For example, performing "DELETE 2" on {1, 2, 3, 4, 5} results in {1, 3, 4, 5}
6. MIN x y: query the participant what is the minimum number in sub-sequence {A_x ... A_y}. 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

 

题意：给定一个序列，每次执行一个操作，对于每个min输出即可。

题解：

好个码农题。。。

Splay处理一下区间翻转，区间加上k，区间左右移动(循环序列)，区间查询。。。

记得开long long。。。。。。

代码:

 #include<bits/stdc++.h>
 using namespace std;
 #define MAXN 100010
 #define MAXM 100010
 #define INF 1e9
 #define LL long long
 struct node
 {
     LL left,right,mn,val,size;
 }tree[MAXN+MAXM];
 LL a[MAXN+MAXM],father[MAXN+MAXM],rev[MAXN+MAXM],tag[MAXN+MAXM];
 LL read()
 {
     LL s=,fh=;char ch=getchar();
     while(ch<''||ch>''){if(ch=='-')fh=-;ch=getchar();}
     while(ch>=''&&ch<=''){s=s*+(ch-'');ch=getchar();}
     return s*fh;
 }
 void Pushup(LL x)
 {
     LL l=tree[x].left,r=tree[x].right;
     tree[x].size=tree[l].size+tree[r].size+;
     tree[x].mn=min(min(tree[l].mn,tree[r].mn),tree[x].val);
 }
 void Build(LL l,LL r,LL f)
 {
     if(l>r)return;
     LL now=l,last=f;
     if(l==r)
     {
         tree[now].val=tree[now].mn=a[l];father[now]=last;
         tree[now].size=;
         if(l<f)tree[last].left=now;
         else tree[last].right=now;
     }
     LL mid=(l+r)/;
     now=mid;
     Build(l,mid-,mid);Build(mid+,r,mid);
     father[now]=last;tree[now].val=a[mid];
     Pushup(now);
     if(mid<f)tree[last].left=now;
     else tree[last].right=now;
 }
 /*void Pushup(int x)
   {
   int l=tree[x].left,r=tree[x].right;
   tree[x].size=tree[l].size+tree[r].size+1;
   tree[x].mn=min(min(tree[l].mn,tree[r].mn),tree[x].val);
   }*/
 void rotate(LL x,LL &root)
 {
     LL y=father[x],z=father[y];
     if(y==root)root=x;
     else
     {
         if(tree[z].left==y)tree[z].left=x;
         else tree[z].right=x;
     }
     if(tree[y].left==x)
     {
         father[x]=z;father[y]=x;tree[y].left=tree[x].right;tree[x].right=y;father[tree[y].left]=y;
     }
     else
     {
         father[x]=z;father[y]=x;tree[y].right=tree[x].left;tree[x].left=y;father[tree[y].right]=y;
     }
     Pushup(y);Pushup(x);
 }
 void Splay(LL x,LL &root)
 {
     while(x!=root)
     {
         int y=father[x],z=father[y];
         if(y!=root)
         {
             if((tree[y].left==x)^(tree[z].left==y))rotate(x,root);
             else rotate(y,root);
         }
         rotate(x,root);
     }
 }
 void Pushdown(LL x)
 {
     LL l=tree[x].left,r=tree[x].right;
     if(tag[x]!=)
     {
         tag[l]+=tag[x];tag[r]+=tag[x];
         tree[l].val+=tag[x];tree[r].val+=tag[x];
         tree[l].mn+=tag[x];tree[r].mn+=tag[x];
         tag[x]=;
     }
     if(rev[x]!=)
     {
         rev[l]^=;rev[r]^=;rev[x]^=;
         swap(tree[x].left,tree[x].right);
     }
 }
 LL Find(LL root,LL rank)
 {
     Pushdown(root);
     if(tree[tree[root].left].size+==rank)return root;
     else if(rank<=tree[tree[root].left].size)return Find(tree[root].left,rank);
     else return Find(tree[root].right,rank-tree[tree[root].left].size-);
 }
 int main()
 {
     LL n,m,i,rt,SIZE,l,r,add,x,y,z,L,R,T,X,P;
     char fh[];
     n=read();
     tree[].val=INF;a[]=tree[].mn=INF;
     tree[].val=INF;tree[n+].val=INF;
     tree[].mn=INF;tree[n+].mn=INF;
     a[]=INF;a[n+]=INF;
     for(i=;i<=n+;i++)a[i]=read(),tree[i].val=tree[i].mn=INF;
     Build(,n+,);
     SIZE=n+;rt=(+n+)/;
     m=read();
     for(i=;i<=m;i++)
     {
         scanf("\n%s",fh);
         if(fh[]=='A')
         {
             l=read();r=read();add=read();
             x=Find(rt,l);y=Find(rt,r+);
             Splay(x,rt);Splay(y,tree[x].right);
             z=tree[y].left;
             tag[z]+=add;tree[z].val+=add;tree[z].mn+=add;
         }
         else if(fh[]=='R')
         {
             if(fh[]=='E')
             {
                 l=read();r=read();
                 x=Find(rt,l);y=Find(rt,r+);
                 Splay(x,rt);Splay(y,tree[x].right);
                 z=tree[y].left;
                 rev[z]^=;
             }
             else
             {
                 l=read();r=read();T=read();
                 L=l;R=r;
                 T=(T%(r-l+)+(r-l+))%(r-l+);
                 if(T==)continue;
                 l=r-T+;
                 x=Find(rt,l);y=Find(rt,r+);
                 Splay(x,rt);Splay(y,tree[x].right);
                 z=tree[y].left;
                 father[z]=;tree[y].left=;
                 Pushup(y);Pushup(x);
                 x=Find(rt,L);y=Find(rt,L+);
                 Splay(x,rt);Splay(y,tree[x].right);
                 father[z]=y;tree[y].left=z;
                 Pushup(y);Pushup(x);
             }
         }
         else if(fh[]=='I')
         {
             X=read();P=read();
             x=Find(rt,X+);y=Find(rt,X+);
             Splay(x,rt);Splay(y,tree[x].right);
             tree[y].left=++SIZE;tree[SIZE].val=P;
             father[SIZE]=y;tree[SIZE].size=;
             tree[SIZE].mn=P;
             Pushup(y);Pushup(x);
         }
         else if(fh[]=='D')
         {
             X=read();
             x=Find(rt,X);y=Find(rt,X+);
             Splay(x,rt);Splay(y,tree[x].right);
             z=tree[y].left;tree[y].left=;
             tree[z].size=;father[z]=;
             //tree[SIZE].val=INF;tree[SIZE].mn=INF;
             Pushup(y);Pushup(x);
         }
         else
         {
             l=read();r=read();
             x=Find(rt,l);y=Find(rt,r+);
             Splay(x,rt);Splay(y,tree[x].right);
             z=tree[y].left;
             printf("%lld\n",tree[z].mn);
         }
     }
     fclose(stdin);
     fclose(stdout);
     return ;
 }