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POJ 3468 A Simple Problem with Integers (Splay)

题目链接:点我~~

题意:给n个数,有两种操作,一种是查询区间和,另一种是在区间上每一个数加上v。

思路:第一次摸splay tree,这题算是个模板题,适合思考人生。。。

//Splay(x,0); 将x变为跟节点

//Splay(x,root); 将x变为root下的节点  维护区间时通常旋转为root的右子树

//所需要维护的区间[l,r],通过旋转后即为 root右孩子的左子树

模板参考链接:ˋ( ° ▽、° )  ̄へ ̄

//#include <bits/stdc++.h>
#include<stdio.h>
#include<string.h>
#include<vector>
#include<iostream>
#include<queue>
#include<cmath>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> PI;
typedef pair< PI, int> PII;
const double eps=1e-5;
const double pi=acos(-1.0);
const int mod=1e9+7;
const int INF=0x3f3f3f3f;
#define Key_value ch[ch[root][1]][0]
const int MAXN = 300000;
const int MAXM = 20100;

int pre[MAXN],ch[MAXN][2],key[MAXN],size[MAXN];
int root,tot1;
LL sum[MAXN];
int s[MAXN],tot2;//内存池和容量
int a[MAXN];
int n,q;
int add[MAXN];

void NewNode(int &r,int father,int k)
{
    if(tot2) r = s[tot2--];//取的时候是tot2--,存的时候就是++tot2
    else r = ++tot1;
    pre[r] = father;
    ch[r][0] = ch[r][1] = 0;
    key[r] = k;
    sum[r] = k;
    size[r] = 1;
}

void Update_add(int r,int v)
{
    if(!r)return;
    add[r] += v;
    key[r] += v;
    sum[r] += (LL)v*size[r];
}
void push_up(int r)
{
    int lson = ch[r][0], rson = ch[r][1];
    size[r] = size[lson] + size[rson] + 1;
    sum[r] = sum[lson] + sum[rson] + key[r];
}
void push_down(int r)
{
    if(add[r])
    {
        int lson = ch[r][0], rson = ch[r][1];
        Update_add(lson,add[r]);
        Update_add(rson,add[r]);
        add[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()
{
    root = tot1 = tot2 = 0;
    ch[root][0] = ch[root][1] = size[root] = pre[root] = 0;
    sum[root] = key[root] = 0;
    NewNode(root,0,-1);
    NewNode(ch[root][1],root,-1);
    for(int i = 1; i <= n; i++)
        scanf("%d",&a[i]);
    Build(Key_value,1,n,ch[root][1]);
    push_up(ch[root][1]);
    push_up(root);
}
//旋转,0为左旋,1为右旋
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);
}
//Splay调整,将r结点调整到goal下面
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)  //得到第k个结点
{
    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);
}

//因为加了个空结点,所以将第l个点旋转到根结点,第r+2个点旋转到根结点的右孩子
//l-1 ~~ r+1
LL Get_Sum(int l,int r)
{
    Splay(Get_kth(root,l),0);
    Splay(Get_kth(root,r+2),root);
    return sum[Key_value];
}

void modify(int l,int r,int v)
{
    Splay(Get_kth(root,l),0);
    Splay(Get_kth(root,r+2),root);
    Update_add(Key_value,v);
    push_up(ch[root][1]);
    push_up(root);
}

void Treavel(int x)
{
    if(x)
    {
        Treavel(ch[x][0]);
        printf("结点:%2d: 左儿子 %2d 右儿子 %2d 父结点 %2d size\
               = %2d\n",x,ch[x][0],ch[x][1],pre[x],size[x]);
        Treavel(ch[x][1]);
    }
}
void debug()
{
    printf("root:%d\n",root);
    Treavel(root);
}

int main()
{
    scanf("%d%d",&n,&q);
    Init();
    char op[20];
    int x,y,z;
    while(q--)
    {
        scanf("%s",op);
        int l,r,v;
        if(op[0]=='Q')
        {
            scanf("%d%d",&l,&r);
            cout<<Get_Sum(l,r)<<endl;;
        }
        else
        {
            scanf("%d%d%d",&l,&r,&v);
            modify(l,r,v);
        }

    }

    return 0;
}

==第二版==

//#include <bits/stdc++.h>
#include<stdio.h>
#include<string.h>
#include<vector>
#include<iostream>
#include<queue>
#include<cmath>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> PI;
typedef pair< PI, int> PII;
const double eps=1e-5;
const double pi=acos(-1.0);
const int mod=1e9+7;
const int INF=0x3f3f3f3f;
#define Key_value ch[ch[root][1]][0]
#define keyTree (ch[ch[root][1]][0])
const int MAXN = 300000;
const int MAXM = 20100;

struct SplayTree
{
    int sz[MAXN];
    int ch[MAXN][2];
    int pre[MAXN];
    int root, top1;
    /*这是题目特定变量*/
    int num[MAXN];
    int val[MAXN];
    int add[MAXN];
    LL sum[MAXN];
    inline void Rotate(int x,int f) //旋转,0为左旋,1为右旋
    {
        int y = pre[x];
        push_down(y);
        push_down(x);
        ch[y][!f] = ch[x][f];
        pre[ ch[x][f] ] = y;
        pre[x] = pre[y];
        if(pre[x]) ch[ pre[y] ][ ch[pre[y]][1] == y ] = x;
        ch[x][f] = y;
        pre[y] = x;
        push_up(y);
    }
    inline void Splay(int x,int goal)  //将x旋转到goal的下面
    {
        push_down(x); //防止pre[x]就是目标点,下面的循环就进不去了,x的信息就传不下去了
        while(pre[x] != goal)
        {
            if(pre[pre[x]] == goal)
            {
                Rotate(x, ch[pre[x]][0] == x);
            }
            else
            {
                int y = pre[x], z = pre[y];
                int f = (ch[z][0] == y);
                if(ch[y][f] == x)
                {
                    Rotate(x, !f), Rotate(x, f);
                }
                else
                {
                    Rotate(y, f), Rotate(x, f);
                }
            }
        }
        push_up(x);
        if(goal == 0) root = x;
    }
    inline void RotateTo(int k,int goal)  //把第k位的数转到goal下边
    {
        int x = root;
        push_down(x);
        while(sz[ ch[x][0] ] != k)
        {
            if(k < sz[ ch[x][0] ])
            {
                x = ch[x][0];
            }
            else
            {
                k -= (sz[ ch[x][0] ] + 1);//多减去1,所以k+1位
                x = ch[x][1];
            }
            push_down(x);
        }
        Splay(x,goal);
    }
    inline int Get_kth(int r,int k)  //得到第k个结点
    {
        push_down(r);
        int t = sz[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);
    }
    //以上一般不修改//////////////////////////////////////////////////////////////////////////////
    void debug()
    {
        printf("%d\n",root);
        Treaval(root);
    }
    void Treaval(int x)
    {
        if(x)
        {
            Treaval(ch[x][0]);
            printf("结点%2d:左儿子 %2d 右儿子 %2d 父结点 %2d size = %2d ,val = %2d\n",x,ch[x][0],ch[x][1],pre[x],sz[x],val[x]);
            Treaval(ch[x][1]);
        }
    }
    //以上Debug


    //以下是题目的特定函数:
    inline void NewNode(int &x,int c)
    {
        x = ++top1;
        ch[x][0] = ch[x][1] = pre[x] = 0;
        sz[x] = 1;

        val[x] = sum[x] = c;/*这是题目特定函数*/
        add[x] = 0;
    }

    //把延迟标记推到孩子  类似线段树
    inline void push_down(int x)  /*这是题目特定函数*/
    {
        if(add[x])
        {
            val[x] += add[x];
            add[ ch[x][0] ] += add[x];
            add[ ch[x][1] ] += add[x];
            sum[ ch[x][0] ] += (LL)sz[ ch[x][0] ] * add[x];
            sum[ ch[x][1] ] += (LL)sz[ ch[x][1] ] * add[x];
            add[x] = 0;
        }
    }
    //把孩子状态更新上来
    inline void push_up(int x)
    {
        sz[x] = 1 + sz[ ch[x][0] ] + sz[ ch[x][1] ];
        /*这是题目特定函数*/
        sum[x] = add[x] + val[x] + sum[ ch[x][0] ] + sum[ ch[x][1] ];
    }

    /*初始化*/
    inline void makeTree(int &x,int l,int r,int f)
    {
        if(l > r) return ;
        int m = (l + r)>>1;
        NewNode(x, num[m]);		/*num[m]权值改成题目所需的*/
        makeTree(ch[x][0], l, m - 1, x);
        makeTree(ch[x][1], m + 1, r, x);
        pre[x] = f;
        push_up(x);
    }
    inline void init(int n)  /*这是题目特定函数*/
    {
        ch[0][0] = ch[0][1] = pre[0] = sz[0] = 0;
        add[0] = sum[0] = 0;
        root = top1 = 0;
        //为了方便处理边界,加两个边界顶点
        NewNode(root, 0);
        NewNode(ch[root][1], 0);
        pre[top1] = root;
        sz[root] = 2;
        for (int i = 0 ; i < n ; i ++) scanf("%d",&num[i]);
        makeTree(keyTree, 0, n-1, ch[root][1]);
        push_up(ch[root][1]);
        push_up(root);
    }
    inline void update( )  /*这是题目特定函数*/
    {
        int l, r, c;
        scanf("%d%d%d",&l,&r,&c);
        RotateTo(l-1,0);
        RotateTo(r+1,root);
        add[keyTree] += c;    //sz当前节点儿子的数量
        sum[keyTree] += (LL)c * sz[ keyTree ];
    }
    inline void query()  /*这是题目特定函数*/
    {
        int l, r;
        scanf("%d%d",&l,&r);
        RotateTo(l-1, 0);
        RotateTo(r+1, root);
        printf("%lld\n",sum[keyTree]);
    }
} spt;

int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    spt.init(n);
    //spt.debug();
    while(m--)
    {
        char op[5];
        scanf("%s",op);
        if(op[0] == 'Q')
        {
            spt.query();
        }
        else
        {
            spt.update();
            //spt.debug();
        }
    }
    return 0;
}

 

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