POJ3468 A Simple Problem with Integers（SplayTree做法）

Description

You have N integers, A1, A2, ... , AN. 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, ... , AN. -1000000000 ≤ Ai ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of Aa, Aa+1, ... , Ab. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of Aa, Aa+1, ... , Ab.

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

这道题真的是用什么树都可以做，伸展树的每个节点维护add,sum,key信息,分别表示以该节点为根节点的整棵树的增加值,和,以及本节点的值.注意add标记在x号节点上时,x节点的sum和key就立即更新,这个add只是表示x节点的儿子节点们没有更新.

AC代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
#include<stdlib.h>
#include<queue>
#include<map>
#include<iomanip>
#include<math.h>
#include<sstream>
using namespace std;
typedef long long ll;
typedef double ld;
using namespace std;
const int maxn=100000+100;
int a[maxn];
int n,q;
struct SplayTree
{
    #define Key_Value ch[ch[root][1]][0]
    int ch[maxn][2],pre[maxn],size[maxn],root,tot1;
    int key[maxn];
    int add[maxn];
    ll sum[maxn];
    void NewNode(int &r,int fa,int k)
    {
        r =++tot1;
        pre[r]=fa;
        add[r]=sum[r]=ch[r][0]=ch[r][1]=0;
        size[r]=1;
        key[r]=k;
    }
    void Update_Add(int r,int ADD)
    {
        if(r==0)return ;
        key[r]+=ADD;
        add[r]+=ADD;
        sum[r]+=(long long)ADD*size[r];
    }
    void Push_Up(int r)
    {
        size[r]=1+size[ch[r][0]]+size[ch[r][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;
        }
    }
    void Build(int &x,int l,int r,int fa)
    {
        if(l>r) return ;
        int mid=(l+r)>>1;
        NewNode(x,fa,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=0;
        ch[root][0]=ch[root][1]=pre[root]=size[root]=key[root]=sum[root]=add[root]=0;
        NewNode(root,0,-1);
        NewNode(ch[root][1],root,-1);
        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)
                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;
    }
    int Get_Kth(int r,int k)
    {
        Push_Down(r);
        int t=size[ch[r][0]]+1;
        if(k==t) return r;
        else if(k<t) return Get_Kth(ch[r][0],k);
        else return Get_Kth(ch[r][1],k-t);
    }
    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);
    }
    long long Query(int l,int r)
    {
        Splay(Get_Kth(root,l),0);
        Splay(Get_Kth(root,r+2),root);
        return sum[Key_Value];
    }
}st;
int main()
{
    while(scanf("%d%d",&n,&q)==2)
    {
        st.init();
        while(q--)
        {
            char op[20];
            int x,y,z;
            scanf("%s",op);
            if(op[0]=='C')
            {
                scanf("%d%d%d",&x,&y,&z);
                st.ADD(x,y,z);
            }
            else if(op[0]=='Q')
            {
                scanf("%d%d",&x,&y);
                printf("%I64d\n",st.Query(x,y));
            }
        }
    }
    return 0;
}