HDU - 4348 To the moon （可持久化线段树，区间查询加累加）

Background 

To The Moon is a independent game released in November 2011, it is a role-playing adventure game powered by RPG Maker. 

The premise of To The Moon is based around a technology that allows us to permanently reconstruct the memory on dying man. In this problem, we'll give you a chance, to implement the logic behind the scene. 

You‘ve been given N integers A [1], A [2],..., A 
. On these integers, you need to implement the following operations: 
1. C l r d: Adding a constant d for every {A i | l <= i <= r}, and increase the time stamp by 1, this is the only operation that will cause the time stamp increase. 
2. Q l r: Querying the current sum of {A i | l <= i <= r}. 
3. H l r t: Querying a history sum of {A i | l <= i <= r} in time t. 
4. B t: Back to time t. And once you decide return to a past, you can never be access to a forward edition anymore. 

.. N, M ≤ 10 5, |A [i]| ≤ 10 9, 1 ≤ l ≤ r ≤ N, |d| ≤ 10 4 .. the system start from time 0, and the first modification is in time 1, t ≥ 0, and won't introduce you to a future state.

Inputn m 

A 1 A 2 ... A n 

... (here following the m operations. )
Output... (for each query, simply print the result. )
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

2 4
0 0
C 1 1 1
C 2 2 -1
Q 1 2
H 1 2 1

Sample Output

4
55
9
15

0
1

题解：简单的 可持久化线段树+区间查询，区间累加。搞了半天才搞懂并调试好所有代码。又一模板get！这里少了下推函数，可持久化线段树的特性，很巧妙。

#include<iostream>
#include<cstdio>
#include<cstring>
#define ll long long
using namespace std;

int lc[3000010];//左孩子
int rc[3000010];//右孩子
ll sum[3000010];//区间和
ll lazy[3000010];//懒惰标记
int n,m,cnt,x,y,z,now;

//原数组，第N次操作的根节点
int A[100010],T[100010];
char op[3];

//建一棵空树
void build(int &now,int l,int r){

now=++cnt;
if(l==r){
sum[now]=A[l];
return;
}
build(lc[now],l,(l+r)/2);
build(rc[now],(l+r)/2+1,r);

//push_up
sum[now]=sum[lc[now]]+sum[rc[now]];
}

void add(int &now,int l,int r,int val,int L,int R){

//新建一个节点，并初始化为前一棵线段树的节点的值
cnt++;
lc[cnt]=lc[now];
rc[cnt]=rc[now];
sum[cnt]=sum[now];
lazy[cnt]=lazy[now];
now=cnt;

sum[now]+=1ll*val*(r-l+1);

if(l==L&&r==R){
lazy[now]+=val;//打上懒惰标记
return;
}
if(r<=(L+R)/2)
add(lc[now],l,r,val,L,(L+R)/2);
else
if(l>(L+R)/2)
add(rc[now],l,r,val,(L+R)/2+1,R);
else{
add(lc[now],l,(L+R)/2,val,L,(L+R)/2);
add(rc[now],(L+R)/2+1,r,val,(L+R)/2+1,R);
}
}

ll query(int now,int l,int r,int L,int R){
if(l==L&&r==R)
return sum[now];

ll tmp=1ll*(r-l+1)*lazy[now];//超级巧妙，不用下推了，直接算就对了！

if(r<=(L+R)/2)
return tmp+query(lc[now],l,r,L,(L+R)/2);

if(l>(L+R)/2)
return tmp+query(rc[now],l,r,(L+R)/2+1,R);

return tmp+query(lc[now],l,(L+R)/2,L,(L+R)/2)+query(rc[now],(L+R)/2+1,r,(L+R)/2+1,R);
}

int main(){

while(~scanf("%d%d",&n,&m)){
now=cnt=0;
for(int i=1;i<=n;++i)
scanf("%d",&A[i]);

build(T[0],1,n);

for(int i=1;i<=m;++i){
scanf("%s",op);
if(op[0]=='C'){
scanf("%d%d%d",&x,&y,&z);
++now;
add(T[now]=T[now-1],x,y,z,1,n);
}
else
if(op[0]=='Q'){
scanf("%d%d",&x,&y);
printf("%I64d\n",query(T[now],x,y,1,n));
}
else
if(op[0]=='H'){
scanf("%d%d%d",&x,&y,&z);
printf("%I64d\n",query(T[z],x,y,1,n));
}
else
scanf("%d",&now);
}
}
}