Splay POJ3468（老題新做）
A Simple Problem with Integers
Time Limit: 5000MS
Memory Limit: 131072KB
64bit IO Format: %I64d
& %I64u
Submit Status Practice POJ
3468
Appoint description:
System Crawler (2014-11-12)
Description
You have N integers, A 1 , A 2 , ... , 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.N numbers, the initial values of A 1 , A 2 , ... , AN . -1000000000 ≤ Ai ≤ 1000000000.Q lines represents an operation.abc " means adding c to each of Aa , Aa +1 , ... , Ab . -10000 ≤ c ≤ 10000.ab " 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
Hint
The sums may exceed the range of 32-bit integers.
這題應該算是線段樹區間入目題目，不過還可以用Splay來做，用Splay來維護序列，用到了平衡二叉樹的一個重要的性質那就是中序遍歷是有序的。人生第一道Splay（感人TAT，QAQ）
代碼如下：
/*************************************************************************
> File Name: Spaly.cpp
> Author: acvcla
> QQ:
> Mail: [email protected]
> Created Time: 2014年11月16日 星期日 00時14分26秒
************************************************************************/
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
typedef long long LL;
const int maxn = 1e5 + 100;
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define pb push_back
LL add[maxn],sum[maxn];
int ch[maxn][2],siz[maxn],key[maxn],pre[maxn],A[maxn];
int root,tot;
void newnode(int &x,int fa,int Key)//新建節點
{
x=++tot;
pre[x]=fa;
siz[x]=1;
key[x]=sum[x]=Key;
ch[x][0]=ch[x][1]=add[x]=0;
}
void Modify(int x,int val){//區間更新
if(!x)return;
add[x]+=val;
key[x]+=val;
sum[x]+=(LL)val*siz[x];
}
void push_down(int x){//下傳標記
if(!add[x])return ;
Modify(ch[x][0],add[x]);
Modify(ch[x][1],add[x]);
add[x]=0;
}
void push_up(int x){//更新節點
siz[x]=siz[ch[x][0]]+siz[ch[x][1]]+1;
sum[x]=sum[ch[x][0]]+sum[ch[x][1]]+key[x];
}
void built(int &x,int L,int R,int fa){
if(L>R)return;
int M=(L+R)>>1;
newnode(x,fa,A[M]);
built(ch[x][0],L,M-1,x);
built(ch[x][1],M+1,R,x);
push_up(x);
}
void Init(int n)//初始化Spaly，添加了兩個虛擬節點，便於提取區間，避免討論
{
root=tot=0;
newnode(root,0,0);
newnode(ch[root][1],root,0);
for(int i=1;i<=n;i++)scanf("%d",A+i);
built(ch[ch[root][1]][0],1,n,ch[root][1]);
push_up(ch[root][1]);
push_up(root);
}
void print(int x){
if(!x)return;
print(ch[x][0]);
printf("%d ",key[x]);
print(ch[x][1]);
}
void Rotate(int x,bool kind){//旋轉，true右旋，false左旋
int y=pre[x];
push_down(y);//下傳標記
push_down(x);
ch[y][!kind]=ch[x][kind];
pre[ch[x][kind]]=y;
ch[x][kind]=y;
if(pre[y])ch[pre[y]][ch[pre[y]][1]==y]=x;//若y的父節點存在將其孩子指針指向x
pre[x]=pre[y];
pre[y]=x;
push_up(y);//更新回來，需要注意的是，要先更新孩子
push_up(x);
}
void Spaly(int x,int goal){//伸展操作，將x旋轉到goal下面
push_down(x);
while(pre[x]!=goal){
if(pre[pre[x]]==goal)Rotate(x,ch[pre[x]][0]==x);
else{
int y=pre[x];
bool kind=(ch[pre[y]][0]==y);
if(ch[y][kind]==x){
Rotate(x,!kind);
Rotate(x,kind);
}else{
Rotate(y,kind);
Rotate(x,kind);
}
}
}
push_up(x);
if(goal==0)root=x;//如果goal是0說明已經將x旋轉到了根，所以要更新root
}
int Get_kth(int x,int k){//序列中的第k個值
int t=siz[ch[x][0]]+1;
if(t==k)return x;
if(t>k)return Get_kth(ch[x][0],k);
return Get_kth(ch[x][1],k-t);
}
int main(){
ios_base::sync_with_stdio(false);
cin.tie(0);
siz[0]=sum[0]=0;//不存在的節點初始化為0避免討論
int n,q,l,r,x;
scanf("%d%d",&n,&q);
Init(n);
char cmd[5];
while(q--){
scanf("%s%d%d",cmd,&l,&r);
Spaly(Get_kth(root,l),0);
Spaly(Get_kth(root,r+2),root);
if(cmd[0]=='Q'){
printf("%lld\n",sum[ch[ch[root][1]][0]]);
}else{
int Add;
scanf("%d",&Add);
Modify(ch[ch[root][1]][0],Add);
push_up(ch[root][1]);
push_up(root);
}
}
return 0;
}
