POJ 3468 A Simple Problem with Integers (splay tree入门)

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 47944		Accepted: 14122
Case Time Limit: 2000MS

Description

You have N integers, A₁, A₂, ... , A_N. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval.

Input

The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000.
The second line contains N numbers, the initial values of A₁, A₂, ... , A_N. -1000000000 ≤ A_i ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of A_a, A_a+1, ... , A_b. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of A_a, A_a+1, ... , A_b.

Output

You need to answer all Q commands in order. One answer in a line.

Sample Input

10 5
1 2 3 4 5 6 7 8 9 10
Q 4 4
Q 1 10
Q 2 4
C 3 6 3
Q 2 4

Sample Output

4
55
9
15

Hint

The sums may exceed the range of 32-bit integers.

Source

POJ Monthly--2007.11.25, Yang Yi

 

 

 

 

 

 

 

 

/* ***********************************************
Author        :kuangbin
Created Time  :2013/8/24 22:13:15
File Name     :F:\2013ACM练习\专题学习\splay_tree_2\POJ3468.cpp
************************************************ */

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
//普通的线段树操作，区间增加一个值，查询区间的和
#define Key_value ch[ch[root][1]][0]
const int MAXN = 100010;
int pre[MAXN],ch[MAXN][2],root,tot1;
int size[MAXN];//子树的结点数
int add[MAXN];//增量的延迟标记
int key[MAXN];
long long sum[MAXN];//子树的和
int s[MAXN],tot2;//内存池和容量
int a[MAXN];//初始的数组，建树的时候用，下标从1开始

//debug部分**********************************
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 add = %2d sum = %I64d\n",x,ch[x][0],ch[x][1],pre[x],size[x],add[x],sum[x]);
        Treavel(ch[x][1]);
    }
}
void debug()
{
    printf("root:%d\n",root);
    Treavel(root);
}
//以上是debug部分**************************************

void NewNode(int &r,int father,int k)
{
    if(tot2)r = tot2--;//取的时候是tot2--,存的时候就是++tot2
    else r = ++tot1;
    pre[r] = father;
    size[r] = 1;
    add[r] = 0;
    key[r] = k;
    sum[r] = k;
    ch[r][0] = ch[r][1] = 0;
}
//给r为根的子树增加值，把当前结点更新掉，加延迟标记
void Update_Add(int r,int ADD)
{
    if(r == 0)return;
    add[r] += ADD;
    key[r] += ADD;
    sum[r] += (long long)ADD*size[r];
}
void push_up(int r)
{
    size[r] = size[ch[r][0]] + size[ch[r][1]] + 1;
    sum[r] = sum[ch[r][0]] + sum[ch[r][1]] + key[r];
}
void push_down(int r)
{
    if(add[r])
    {
        Update_Add(ch[r][0],add[r]);
        Update_Add(ch[r][1],add[r]);
        add[r] = 0;
    }
}
//建树
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);
}
int n,q;
//初始化
void Init()
{
    root = tot1 = tot2 = 0;
    ch[root][0] = ch[root][1] = pre[root] = size[root] = add[root] = sum[root] = key[root] = 0;
    //加两个虚结点
    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);//先把y的标记下传，在把x的标记下传
    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)
            Rotate(r,ch[pre[r]][0]==r);
        else
        {
            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;
}
//得到第k个结点
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);
}

//区间增加一个值
//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
//那么Key_value就是需要修改的区间[l,r]
void ADD(int l,int r,int D)
{
    Splay(Get_kth(root,l),0);
    Splay(Get_kth(root,r+2),root);
    Update_Add(Key_value,D);
    push_up(ch[root][1]);
    push_up(root);
}
//查询区间的和
long long Query_sum(int l,int r)
{
    Splay(Get_kth(root,l),0);
    Splay(Get_kth(root,r+2),root);
    return sum[Key_value];
}

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    while(scanf("%d%d",&n,&q) == 2)
    {
        Init();
        //debug();
        int x,y,z;
        char op[20];
        while(q--)
        {
            scanf("%s",op);
            if(op[0] == 'Q')
            {
                scanf("%d%d",&x,&y);
                printf("%I64d\n",Query_sum(x,y));
            }
            else
            {
                scanf("%d%d%d",&x,&y,&z);
                ADD(x,y,z);
            }
        }
    }
    return 0;
}