POJ 3580

题目链接：http://poj.org/problem?id=3580

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

题意：

给出五种操作：

  1. ADD x y D: 对区间[x,y]全部加上D；
  2. REVERSE x y: 反转区间[x,y]；
  3. REVOLVE x y T: 将区间[x,y]向右循环平移T次，每次一格，例如 REVOLVE 2 4 2 在{1, 2, 3, 4, 5}操作结果为{1, 3, 4, 2, 5}；
  4. INSERT x P: 在位置x后面插入一个数P；
  5. DELETE x: 删除位置x上的数；
  6. MIN x y: 求区间[x,y]中的最小值；

题解：

splay模板题，lazy标记模仿线段树标记区间加上多少，rev标记标记是否需要反转，mini[x]记录节点x统领的整棵子树（包含自己）的key[x]的最小值。

AC代码：

#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn=2e5+10;
int n,m;
int a[maxn];

/******************************** splay - st ********************************/
#define Key_value ch[ch[root][1]][0]
int root,nodecnt;
int par[maxn],ch[maxn][2];
int key[maxn],mini[maxn],size[maxn];
int lazy[maxn]; //lazy标记
bool rev[maxn]; //反转标记
int pool[maxn],poolsize; //节点回收
void NewNode(int &x,int p,int k)
{
    if(poolsize>0) x=pool[--poolsize];
    else x=++nodecnt;
    par[x]=p;
    ch[x][0]=ch[x][1]=0;
    key[x]=k;
    mini[x]=k;
    size[x]=1;
    lazy[x]=0;
    rev[x]=0;
}
void Update_Rev(int x)
{
    if(x==0) return;
    swap(ch[x][0],ch[x][1]);
    rev[x]^=1;
}
void Update_Add(int x,int val)
{
    if(x==0) return;
    key[x]+=val;
    mini[x]+=val;
    lazy[x]+=val;
}
void Pushup(int x)
{
    size[x]=size[ch[x][0]]+size[ch[x][1]]+1;

    mini[x]=key[x];
    if(ch[x][0]) mini[x]=min(mini[x],mini[ch[x][0]]);
    if(ch[x][1]) mini[x]=min(mini[x],mini[ch[x][1]]);
}
void Pushdown(int x)
{
    if(rev[x])
    {
        Update_Rev(ch[x][0]);
        Update_Rev(ch[x][1]);
        rev[x]=0;
    }
    if(lazy[x])
    {
        Update_Add(ch[x][0],lazy[x]);
        Update_Add(ch[x][1],lazy[x]);
        lazy[x]=0;
    }
}
void Inorder(int x) //debug
{
    if(x==0) return;
    Pushdown(x);
    Inorder(ch[x][0]);
    printf("%d ",key[x]);
    Inorder(ch[x][1]);
    Pushup(x);
    if(x==root) printf("
");
}
void Rotate(int x,int type) //旋转，0为左旋zag，1为右旋zig
{
    int y=par[x];
    Pushdown(y); Pushdown(x); //先把y的标记向下传递,再把x的标记往下传递
    ch[y][!type]=ch[x][type]; par[ch[x][type]]=y;
    if(par[y]) ch[par[y]][(ch[par[y]][1]==y)]=x;
    par[x]=par[y];
    ch[x][type]=y; par[y]=x;
    Pushup(y); Pushup(x);
}
void Splay(int x,int goal)
{
    while(par[x]!=goal)
    {
        if(par[par[x]]==goal) Rotate(x,ch[par[x]][0]==x); //左孩子zig，右孩子zag
        else
        {
            Pushdown(par[par[x]]); Pushdown(par[x]); Pushdown(x);
            int y=par[x];
            int type=(ch[par[y]][0]==y); //type=0，y是右孩子；type=1，y是左孩子
            if(ch[y][type]==x)
            {
                Rotate(x,!type);
                Rotate(x,type);
            }
            else
            {
                Rotate(y,type);
                Rotate(x,type);
            }
        }
    }
    if(goal==0) root=x;
}
int Get_Kth(int x,int k) //得到第k个节点
{
    Pushdown(x);
    int t=size[ch[x][0]]+1;
    if(t==k) return x;
    if(t>k) return Get_Kth(ch[x][0],k);
    else return Get_Kth(ch[x][1],k-t);
}
int Get_Min(int x)
{
    Pushdown(x);
    while(ch[x][0])
    {
        x=ch[x][0];
        Pushdown(x);
    }
    return x;
}
int Get_Max(int x)
{
    Pushdown(x);
    while(ch[x][1])
    {
        x=ch[x][1];
        Pushdown(x);
    }
    return x;
}
void Build(int &x,int l,int r,int par) //建树，先建立中间结点，再建两端的方法
{
    if(l>r) return;
    int mid=(l+r)/2;
    NewNode(x,par,a[mid]);
    Build(ch[x][0],l,mid-1,x);
    Build(ch[x][1],mid+1,r,x);
    Pushup(x);
}
void Init() //初始化，前后各加一个空节点
{
    root=nodecnt=poolsize=0;
    par[root]=ch[root][0]=ch[root][1]=0;
    key[root]=size[root]=0;
    lazy[root]=rev[root]=0;
    NewNode(root,0,-1); //头部加入一个空位
    NewNode(ch[root][1],root,-1); //尾部加入一个空位
    Build(Key_value,1,n,ch[root][1]);
    Pushup(ch[root][1]);
    Pushup(root);
}
void Add(int l,int r,int val)
{
    Splay(Get_Kth(root,l-1+1),0); //l的前驱l-1伸展到根
    Splay(Get_Kth(root,r+1+1),root); //r的后继r+1伸展到根的右孩子
    Update_Add(Key_value,val);
    Pushup(ch[root][1]);
    Pushup(root);
}
void Move(int l,int r,int p) //截取[l,r]放到位置p之后
{
    Splay(Get_Kth(root,l-1+1),0); //l的前驱l-1伸展到根
    Splay(Get_Kth(root,r+1+1),root); //r的后继r+1伸展到根的右孩子
    int tmp=Key_value; //Key_value=ch[ch[root][1]][0]所统领的子树即[l,r]
    Key_value=0; //剥离[l,r]子树
    Pushup(ch[root][1]); Pushup(root);
    Splay(Get_Kth(root,p+0+1),0); //p伸展到根
    Splay(Get_Kth(root,p+1+1),root); //p的后继p+1伸展到根的右孩子
    Key_value=tmp; par[Key_value]=ch[root][1]; //接上[l,r]子树
    Pushup(ch[root][1]); Pushup(root);
}
void Reverse(int l,int r) //反转[l,r]区间
{
    Splay(Get_Kth(root,l-1+1),0);
    Splay(Get_Kth(root,r+1+1),root);
    Update_Rev(Key_value);
    Pushup(ch[root][1]);
    Pushup(root);
}
void Revolve(int l,int r,int T)
{
    int D=r-l+1;
    int L=T%D;
    if(L==0) return;
    Move(r-L+1,r,l-1);
}
void Insert(int p,int k)
{
    Splay(Get_Kth(root,p+0+1),0); //p伸展到根
    Splay(Get_Kth(root,p+1+1),root); //p的后继p+1伸展到根的右孩子
    Inorder(Key_value);
    NewNode(Key_value,ch[root][1],k);
    Pushup(ch[root][1]); Pushup(root);
}
void Collect(int x) //回收节点x统领的子树
{
    if(x==0) return;
    pool[poolsize++]=x;
    Collect(ch[x][0]);
    Collect(ch[x][1]);
}
void Delete(int p)
{
    Splay(Get_Kth(root,p-1+1),0); //p的前驱p-1伸展到根
    Splay(Get_Kth(root,p+1+1),root); //p的后继p+1伸展到根的右孩子
    Collect(Key_value);
    par[Key_value]=0;
    Key_value=0;
    Pushup(ch[root][1]); Pushup(root);
}
int Min(int l,int r)
{
    Splay(Get_Kth(root,l-1+1),0);
    Splay(Get_Kth(root,r+1+1),root);
    return mini[Key_value];
}
/******************************** splay - ed ********************************/

int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++) scanf("%d",&a[i]);
    Init();

    scanf("%d",&m);
    char op[50];
    for(int i=1;i<=m;i++)
    {
        scanf("%s",op);
        if(op[0]=='A')
        {
            int x,y,D; scanf("%d%d%d",&x,&y,&D);
            Add(x,y,D);
        }
        if(op[0]=='R' && op[3]=='E')
        {
            int x,y; scanf("%d%d",&x,&y);
            Reverse(x,y);
        }
        if(op[0]=='R' && op[3]=='O')
        {
            int x,y,T; scanf("%d%d%d",&x,&y,&T);
            Revolve(x,y,T);
        }
        if(op[0]=='I')
        {
            int x,p; scanf("%d%d",&x,&p);
            Insert(x,p);
        }
        if(op[0]=='D')
        {
            int x; scanf("%d",&x);
            Delete(x);
        }
        if(op[0]=='M')
        {
            int x,y; scanf("%d%d",&x,&y);
            printf("%d
",Min(x,y));
        }
        //Inorder(root);
    }
}

两组数据：

5
1 2 3 4 5 11
ADD 1 4 1
REVERSE 2 4
ADD 2 5 2
REVOLVE 2 4 2
DELETE 4
MIN 1 3
REVERSE 1 4
INSERT 2 4
REVOLVE 1 4 6
DELETE 2
MIN 1 4
10
1 2 3 4 5 6 7 8 9 10
15
ADD 4 8 3
MIN 5 7
MIN 7 10
REVERSE 2 5
MIN 2 6
MIN 2 3
INSERT 3 4
MIN 3 4
MIN 5 10
DELETE 6
MIN 3 5
MIN 4 4
REVOLVE 3 6 7
MIN 5 8
MIN 7 10

第一次写完之后一些BUG通过第一组数据调完了，感觉应该没问题了，然后交了发现WA，

然后用第二组数据一测，发现居然是因为用来debug的一些输出忘记删掉了……尴尬……

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原文地址：https://www.cnblogs.com/dilthey/p/9404613.html