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
依旧是几种基本操作的组合==值得注意的是:
1.由于要求的是区间最小值，初始化的值（没有用的点）就得初始化为inf
2.有一个脑残的错误
Update_Same(int r,int v)
{
if(!r)return;
key[r]+=v;
sum[r]+=v;3.为了以防万一
sum[r]=key[r];
if(ch[r][0]) sum[r]=min(sum[r],sum[ch[r][0]]);
if(ch[r][1]) sum[r]=min(sum[r],sum[ch[r][1]]);#include <iostream>
#include <string.h>
#include <stdio.h>
#include <algorithm>
using namespace std;
#define Key_value ch[ch[root][1]][0]
const int MAXN=200010;
const int INF=0x3f3f3f3f;
int pre[MAXN],ch[MAXN][2],key[MAXN],size[MAXN];
int sum[MAXN],rev[MAXN],same[MAXN];
int lx[MAXN],rx[MAXN],mx[MAXN];
int root,tot1;
int s[MAXN],tot2;
int a[MAXN];
int n,q;
void NewNode(int &r,int father,int k)
{
if(tot2)r=s[tot2--];
else r=++tot1;
pre[r]=father;
ch[r][0]=ch[r][1]=0;
key[r]=k;
sum[r]=k;
rev[r]=same[r]=0;
// lx[r]=rx[r]=mx[r]=k;
size[r]=1;
}
void Update_Same(int r,int v)
{
if(!r)return;
key[r]+=v;
sum[r]+=v;
// lx[r]=rx[r]=mx[r]=max(v,v*size[r]);
same[r]+=v;
}
void Update_Rev(int r)
{
if(!r)return;
swap(ch[r][0],ch[r][1]);
// swap(lx[r],rx[r]);
rev[r]^=1;//这里要注意，一定是异或1
}
void Push_Up(int r)
{
int lson=ch[r][0],rson=ch[r][1];
size[r]=size[lson]+size[rson]+1;
//-1!!!
sum[r]=key[r];
if(ch[r][0]) sum[r]=min(sum[r],sum[ch[r][0]]);
if(ch[r][1]) sum[r]=min(sum[r],sum[ch[r][1]]);
}
void Push_Down(int r)
{
if(same[r])
{
Update_Same(ch[r][0],same[r]);//!!
Update_Same(ch[r][1],same[r]);
same[r]=0;
}
if(rev[r])
{
Update_Rev(ch[r][0]);
Update_Rev(ch[r][1]);
rev[r]=0;
}
}
void Build(int &x,int l,int r,int father)
{
if(l>r)return;
int mid=(l+r)/2;
NewNode(x,father,a[mid]);
Build(ch[x][0],l,mid-1,x);
Build(ch[x][1],mid+1,r,x);
Push_Up(x);
}
void Init()
{
for(int i=1;i<=n;i++)scanf("%d",&a[i]);
root=tot1=tot2=0;
ch[root][0]=ch[root][1]=pre[root]=size[root]=same[root]=0;
sum[root]=INF;
// key[root]=0;
NewNode(root,0,INF);
NewNode(ch[root][1],root,INF);//头尾各加入一个空位
Build(Key_value,1,n,ch[root][1]);
Push_Up(ch[root][1]);
Push_Up(root);
}
void Rotate(int x,int kind)
{
int y=pre[x];
Push_Down(y);
Push_Down(x);
ch[y][!kind]=ch[x][kind];
pre[ch[x][kind]]=y;
if(pre[y])
ch[pre[y]][ch[pre[y]][1]==y]=x;
pre[x]=pre[y];
ch[x][kind]=y;
pre[y]=x;
Push_Up(y);
}
void Splay(int r,int goal)
{
Push_Down(r);
while(pre[r]!=goal)
{
if(pre[pre[r]]==goal)
{
Push_Down(pre[r]);
Push_Down(r);
Rotate(r,ch[pre[r]][0]==r);
}
else
{
Push_Down(pre[pre[r]]);
Push_Down(pre[r]);
Push_Down(r);
int y=pre[r];
int kind=ch[pre[y]][0]==y;
if(ch[y][kind]==r)
{
Rotate(r,!kind);
Rotate(r,kind);
}
else
{
Rotate(y,kind);
Rotate(r,kind);
}
}
}
Push_Up(r);
if(goal==0)root=r;
}
int Get_Kth(int r,int k)
{
Push_Down(r);
int t=size[ch[r][0]]+1;
if(t==k)return r;
if(t>k)return Get_Kth(ch[r][0],k);
else return Get_Kth(ch[r][1],k-t);
}
//在第pos个数后插入tot个数
void Insert(int pos,int tot,int tmp)
{
// for(int i=0;i<tot;i++)a[i]=tmp;
Splay(Get_Kth(root,pos+1),0);
Splay(Get_Kth(root,pos+2),root);
// Build(Key_value,0,tot-1,ch[root][1]);
NewNode(Key_value,ch[root][1],tmp);
Push_Up(ch[root][1]);
Push_Up(root);
}
void erase(int r)
{
if(r)
{
s[++tot2]=r;
erase(ch[r][0]);
erase(ch[r][1]);
}
}
//从第pos个数开始连续删除tot个数
void Delete(int pos,int tot)
{
Splay(Get_Kth(root,pos),0);
Splay(Get_Kth(root,pos+tot+1),root);
erase(Key_value);
// s[++tot2]=Key_value;
pre[Key_value]=0;
Key_value=0;
Push_Up(ch[root][1]);
Push_Up(root);
}
//从第pos个数连续开始的tot个数修改为c=>增加c
void Make_Same(int pos,int tot,int c)
{
Splay(Get_Kth(root,pos),0);
Splay(Get_Kth(root,pos+tot+1),root);
Update_Same(Key_value,c);
Push_Up(ch[root][1]);
Push_Up(root);
}
//反转
void Reverse(int pos,int tot)
{
Splay(Get_Kth(root,pos),0);
Splay(Get_Kth(root,pos+tot+1),root);
Update_Rev(Key_value);
Push_Up(ch[root][1]);
Push_Up(root);
}
//求和=>求最小值
int Get_Sum(int pos,int tot)
{
Splay(Get_Kth(root,pos),0);
Splay(Get_Kth(root,pos+tot+1),root);
return sum[Key_value];
}
void revolve(int l,int r,int T)
{
// Splay(Get_Kth(root,pos),0);
// Splay(Get_Kth(root,pos+tot+1),root);
// int tmp=Key_value;
// Key_value=0;
// Push_Up(ch[root][1]);
// Push_Up(root);
// Splay(Get_Kth(root,pos1+1),0);
// Splay(Get_Kth(root,pos1+2),root);
// Key_value=tmp;
// pre[Key_value]=ch[root][1];
// Push_Up(ch[root][1]);
// Push_Up(root);
int len=r-l+1;
T=(T%len+len)%len;
if(T==0)return;
int c=r-T+1;//将区间[c,r]放在[l,c-1]前面
Splay(Get_Kth(root,c),0);
Splay(Get_Kth(root,r+2),root);
int tmp=Key_value;
Key_value=0;
Push_Up(ch[root][1]);
Push_Up(root);
Splay(Get_Kth(root,l),0);
Splay(Get_Kth(root,l+1),root);
Key_value=tmp;
pre[Key_value]=ch[root][1];//这个不用忘记
Push_Up(ch[root][1]);
Push_Up(root);
}
void Inorder(int r)
{
if(!r)return;
Push_Down(r);
Inorder(ch[r][0]);
printf("%d ",sum[r]);
Inorder(ch[r][1]);
}
int main()
{
// freopen("cin.txt","r",stdin);
//freopen("out.txt","w",stdout);
while(~scanf("%d",&n))
{
Init();
char op[20];
int x,y,z;
scanf("%d",&q);
while(q--)
{
scanf("%s",op);
if(strcmp(op,"INSERT")==0)
{
scanf("%d%d",&x,&y);
Insert(x,1,y);
}
else if(strcmp(op,"DELETE")==0)
{
scanf("%d%d",&x);
Delete(x,1);
}
else if(strcmp(op,"ADD")==0)
{
scanf("%d%d%d",&x,&y,&z);
Make_Same(x,y-x+1,z);
}
else if(strcmp(op,"REVERSE")==0)
{
scanf("%d%d",&x,&y);
Reverse(x,y-x+1);
}
else if(op[0]=='M')
{
scanf("%d%d",&x,&y);
printf("%d\n",Get_Sum(x,y-x+1));
}
else if(strcmp(op,"REVOLVE")==0)
{
scanf("%d%d%d",&x,&y,&z);
// z=z%(y-x+1);
// if(z==0) continue;
int a=y-z+1,b=y,c=x;
revolve(x,y,z);
}
}
//Inorder(root);
}
return 0;
}
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