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POJ 3580(SuperMemo-Splay区间加)[template:Splay V2]

2024-09-21 13:05:45   217人阅读

SuperMemo
Time Limit: 5000MS   Memory Limit: 65536K
Total Submissions: 11384   Accepted: 3572
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:

  1. 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}
  2. 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}
  3. 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}
  4. 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}
  5. DELETE x: delete Ax. 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 {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


Splay裸题

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#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (200000+10)
#define MAXM (100000+10) 
typedef long long ll;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+(a-b)/F*F+F)%F;}
int modF(int a,int b){return (a+a/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}

int n,m;
int a[MAXN];
class Splay
{
public:
	int father[MAXN],siz[MAXN],n;
	int ch[MAXN][2],val[MAXN];
	bool root[MAXN],rev[MAXN];	
	int addv[MAXN],minv[MAXN];
	int roo; //root
	void mem(int _n)
	{
		MEM(father) MEM(siz) MEM(root) MEM(rev)	MEM(ch) MEM(val) flag=0; MEM(addv) MEM(minv)
		n=0; 
		roo=1; 
		build(roo,1,_n,0);root[1]=1;
	}
	void newnode(int &x,int f,int v)
	{
		x=++n;
		father[x]=f;
		val[x]=minv[x]=v;siz[x]=1;
	}
	
	void build(int &x,int L,int R,int f)
	{
		if (L>R) return ;
		int m=(L+R)>>1;
		newnode(x,f,a[m]); 
		build(ch[x][0],L,m-1,x);
		build(ch[x][1],m+1,R,x);
		maintain(x);
	}
	int getkth(int x,int k)
	{
		pushdown(x); 
		int t;
		if (ch[x][0]) t=siz[ch[x][0]]; else t=0;
		
		if (t==k-1) return x;
		else if (t>=k) return getkth(ch[x][0],k);
		else return getkth(ch[x][1],k-t-1);
		
	}
	
	
	void pushdown(int x)
	{
		if (x) if (rev[x])
		{
			swap(ch[x][0],ch[x][1]);
			if (ch[x][0]) rev[ ch[x][0] ]^=1;
			if (ch[x][1]) rev[ ch[x][1] ]^=1;
			rev[x]^=1;
		}
		if (addv[x])
		{
			if (ch[x][0]) addv[ ch[x][0] ]+=addv[x],minv[ ch[x][0] ]+=addv[x],val[ ch[x][0] ]+=addv[x];
			if (ch[x][1]) addv[ ch[x][1] ]+=addv[x],minv[ ch[x][1] ]+=addv[x],val[ ch[x][1] ]+=addv[x];
			addv[x]=0;
		}
	}
	void maintain(int x)
	{
		siz[x]=siz[ch[x][0]]+siz[ch[x][1]]+1;
		minv[x]=val[x];
		if (ch[x][0]) minv[ x ]=min(minv[x],minv[ ch[x][0] ] + addv[x] );
		if (ch[x][1]) minv[ x ]=min(minv[x],minv[ ch[x][1] ] + addv[x] );
	}
	void rotate(int x)
	{
		int y=father[x],kind=ch[y][1]==x;
		
		pushdown(y); pushdown(x);
			
		ch[y][kind]=ch[x][!kind];
		if (ch[y][kind]) {
			father[ch[y][kind]]=y;
		}
		father[x]=father[y];
		father[y]=x;
		ch[x][!kind]=y;
		if (root[y])
		{
			root[x]=1;root[y]=0;roo=x;
		}
		else
		{
			ch[father[x]][ ch[father[x]][1]==y ] = x;
		}
		maintain(y);maintain(x);
	}
	void splay(int x)
	{
		while(!root[x])
		{
			int y=father[x];
			int z=father[y];
			if (root[y]) rotate(x);
			else if ( (ch[y][1]==x)^(ch[z][1]==y) )
			{
				rotate(x); rotate(x);
			} 
			else 
			{
				rotate(y); rotate(x);
			}
		}
		roo=x;
	}
	void splay(int x,int r)
	{
		while(!(father[x]==r))
		{
			int y=father[x];
			int z=father[y];
			if (father[y]==r) rotate(x);
			else if ( (ch[y][1]==x)^(ch[z][1]==y) )
			{
				rotate(x); rotate(x);
			} 
			else 
			{
				rotate(y); rotate(x);
			}
		}
	}
	
	void Cut(int a,int b,int c)
	{
		int x=getkth(roo,a),y=getkth(roo,b);
		splay(x);
		splay(y,roo);
		pushdown(x);pushdown(y);
		int z=ch[y][0];
		ch[y][0]=0; maintain(y); maintain(x);
		
		int u=getkth(roo,c),v=getkth(roo,c+1);
		splay(u);
		splay(v,roo);
		pushdown(u);pushdown(v);
		ch[v][0]=z;father[z]=v;
		maintain(v);maintain(u);
		
	}
	
	void Flip(int a,int b)
	{
		int x=getkth(roo,a),y=getkth(roo,b);
		splay(x);
		splay(y,roo);
		pushdown(x);pushdown(y);
		int z=ch[y][0];
		rev[z]^=1;
		maintain(y); maintain(x);
	} 
	
	void Add(int a,int b,int c)
	{
		int x=getkth(roo,a),y=getkth(roo,b);
		splay(x);
		splay(y,roo);
		pushdown(x);pushdown(y);
		int z=ch[y][0];
		addv[z]+=c; val[z]+=c; minv[z]+=c;
		maintain(y); maintain(x);
	} 
	
	int queryMin(int a,int b)
	{
		int x=getkth(roo,a),y=getkth(roo,b);
		splay(x);
		splay(y,roo);
		pushdown(x);pushdown(y);
		int z=ch[y][0];		
		maintain(y); maintain(x);
		return minv[z];
	} 
	
	void insert(int a,int P)
	{
		int x=getkth(roo,a),y=getkth(roo,a+1);
		splay(x);
		splay(y,roo);
		pushdown(x);pushdown(y);
		newnode(ch[y][0],y,P);
		maintain(y); maintain(x);
	}
	void Delete(int a,int b)
	{
		int x=getkth(roo,a),y=getkth(roo,b);
		splay(x);
		splay(y,roo);
		pushdown(x);pushdown(y);
		int z=ch[y][0];
		ch[y][0]=0; father[z]=0; maintain(y); maintain(x);
		
	}
	
	
	bool flag;
	void print(int x)
	{
		if (x==0) return ;
		pushdown(x);
		print(ch[x][0]);
		printf("%d ",val[x]);
		print(ch[x][1]);	
	} 
	
}S;

char s[20];

int main()
{
//	freopen("poj3580.in","r",stdin);
//	freopen(".out","w",stdout);
	
	while(cin>>n)
	{
		For(i,n) scanf("%d",&a[i+1]); a[1]=a[n+2]=INF;
		S.mem(n+2);
		cin>>m; 
		For(i,m)
		{
			scanf("%s",s);
			if (s[0]==‘A‘) //ADD 
			{
				int x,y,D;
				scanf("%d%d%d",&x,&y,&D);
				S.Add(x,y+2,D);
			} else if (s[0]==‘I‘) { //INSERT 
				int x,P;
				scanf("%d%d",&x,&P);
				S.insert(x+1,P);
			} else if (s[0]==‘D‘) { //DELETE 
				int x;
				scanf("%d",&x);
				S.Delete(x,x+2);
			} else if (s[0]==‘M‘) { //MIN
				int x,y;
				scanf("%d%d",&x,&y);
				printf("%d\n",S.queryMin(x,y+2));
			} else if (s[3]==‘E‘) { //REVERSE
				int x,y;
				scanf("%d%d",&x,&y);
				S.Flip(x,y+2);
			} else {  //REVOLVE 
				int x,y,t;
				scanf("%d%d%d",&x,&y,&t);
				t=(t%(y-x+1)+(y-x+1))%(y-x+1);
				if (t) S.Cut(y+2-t-1,y+2,x); 
			}
	//		S.print(S.roo);cout<<endl;		
		}
		
	}
	
	
	return 0;
}


POJ 3580(SuperMemo-Splay区间加)[template:Splay V2]


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