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 = ;
 int pre[MAXN],ch[MAXN][],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][]);
         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][],ch[x][],pre[x],size[x],add[x],sum[x]);
         Treavel(ch[x][]);
     }
 }
 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] = ;
     add[r] = ;
     key[r] = k;
     sum[r] = k;
     ch[r][] = ch[r][] = ;
 }
 //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 == )return;
     add[r] += ADD;
     key[r] += ADD;
     sum[r] += (long long)ADD*size[r];
 }
 void push_up(int r)
 {
     size[r] = size[ch[r][]] + size[ch[r][]] + ;
     sum[r] = sum[ch[r][]] + sum[ch[r][]] + key[r];
 }
 void push_down(int r)
 {
     if(add[r])
     {
         Update_Add(ch[r][],add[r]);
         Update_Add(ch[r][],add[r]);
         add[r] = ;
     }
 }
 //建树
 void Build(int &x,int l,int r,int father)
 {
     if(l > r)return;
     int mid = (l+r)/;
     NewNode(x,father,a[mid]);
     Build(ch[x][],l,mid-,x);
     Build(ch[x][],mid+,r,x);
     push_up(x);
 }
 int n,q;
 //初始化
 void Init()
 {
     root = tot1 = tot2 = ;
     ch[root][] = ch[root][] = pre[root] = size[root] = add[root] = sum[root] = key[root] = ;
     //加两个虚结点
     NewNode(root,,-);
     NewNode(ch[root][],root,-);
     for(int i = ;i <= n;i++)
         scanf("%d",&a[i]);
     Build(Key_value,,n,ch[root][]);
     push_up(ch[root][]);
     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]][]==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]][]==r);
         else
         {
             int y = pre[r];
             int kind = ch[pre[y]][]==y;
             if(ch[y][kind] == r)
             {
                 Rotate(r,!kind);
                 Rotate(r,kind);
             }
             else
             {
                 Rotate(y,kind);
                 Rotate(r,kind);
             }
         }
     }
     push_up(r);
     if(goal == ) root = r;
 }
 //得到第k个结点
 int Get_kth(int r,int k)
 {
     push_down(r);
     int t = size[ch[r][]] + ;
     if(t == k)return r;
     if(t > k)return Get_kth(ch[r][],k);
     else return Get_kth(ch[r][],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),);
     Splay(Get_kth(root,r+),root);
     Update_Add(Key_value,D);
     push_up(ch[root][]);
     push_up(root);
 }
 //查询区间的和
 long long Query_sum(int l,int r)
 {
     Splay(Get_kth(root,l),);
     Splay(Get_kth(root,r+),root);
     return sum[Key_value];
 }

 int main()
 {
     //freopen("in.txt","r",stdin);
     //freopen("out.txt","w",stdout);
     while(scanf("%d%d",&n,&q) == )
     {
         Init();
         //debug();
         int x,y,z;
         char op[];
         while(q--)
         {
             scanf("%s",op);
             if(op[] == '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 ;
 }