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

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 ;

 }

巴特西

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

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