poj3580supermemo【splay】

来源：互联网发布：软件行业的发展编辑：程序博客网时间：2024/06/06 09:48

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:

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}
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}
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}
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}
DELETE x: delete A_x. 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 {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

51 2 3 4 52ADD 2 4 1MIN 4 5

Sample Output

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;

当中sum[r]不能附成key[r]

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]]);

4.右移的操作可以看做是剪切一段挪过去~~就和上个题一样啦~·~

#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cvoid 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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