HDU 4348 To the moon（可持久化线段树）
HDU 4348 To the moon（可持久化线段树）
To the moon
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 4287 Accepted Submission(s): 923
Problem Description
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[N]. On these integers, you need to implement the following operations: 1. C l r d: Adding a constant d for every {Ai | 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 {Ai | l <= i <= r}. 3. H l r t: Querying a history sum of {Ai | 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 ≤ 105, |A[i]| ≤ 109, 1 ≤ l ≤ r ≤ N, |d| ≤ 104 .. 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.
Input
n m A1 A2 ... An ... (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 1Sample Output
4
55
9
15
0
1
可持久化线段树，这道题目会卡内存，所以在pushdown和pushup的时候，新建节点可能会超内存那么我就可以跳过pushup和pushdown#include <iostream>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <stdlib.h>
#include <stdio.h>
using namespace std;
typedef long long int LL;
const int maxn=1e5;
int rt[maxn*35+5];
int ls[maxn*35+5];
int rs[maxn*35+5];
int p;
LL sum[maxn*35+5];
LL pos[maxn*35+5];
int n,m,t;
int newnode()
{
ls[p]=rs[p]=sum[p]=pos[p]=0;
return p++;
}
void build(int &node,int begin,int end)
{
if(!node) node=newnode();
if(begin==end)
{
scanf("%lld",&sum[node]);
return;
}
int mid=(begin+end)>>1;
build(ls[node],begin,mid);
build(rs[node],mid+1,end);
sum[node]=sum[ls[node]]+sum[rs[node]];
}
void update(int &node,int begin,int end,int left,int right,int val)
{
sum[p]=sum[node];ls[p]=ls[node];rs[p]=rs[node];
pos[p]=pos[node];
node=p;p++;
sum[node]+=1LL*val*(right-left+1);
if(left==begin&&end==right)
{
pos[node]+=val;
return;
}
int mid=(begin+end)>>1;
if(right<=mid) update(ls[node],begin,mid,left,right,val);
else if(left>mid) update(rs[node],mid+1,end,left,right,val);
else
{
update(ls[node],begin,mid,left,mid,val);
update(rs[node],mid+1,end,mid+1,right,val);
}
}
LL query(int node,int begin,int end,int left,int right)
{
if(left<=begin&&end<=right) return sum[node];
LL ret=1LL*pos[node]*(right-left+1);
int mid=(begin+end)>>1;
if(right<=mid) ret+=query(ls[node],begin,mid,left,right);
else if(left>mid) ret+=query(rs[node],mid+1,end,left,right);
else
{
ret+=(query(ls[node],begin,mid,left,mid)+query(rs[node],mid+1,end,mid+1,right));
}
return ret;
}
int main()
{
char x;
while(scanf("%d%d",&n,&m)!=EOF)
{
t=0;
p=0;
build(rt[0],1,n);
int l,r,d,time;
LL ans;
for(int i=1;i<=m;i++)
{
cin>>x;
if(x=='C')
{
scanf("%d%d%d",&l,&r,&d);
update(rt[++t]=rt[t-1],1,n,l,r,d);
}
else if(x=='Q')
{
scanf("%d%d",&l,&r);
ans=query(rt[t],1,n,l,r);
printf("%lld\n",ans);
}
else if(x=='H')
{
scanf("%d%d%d",&l,&r,&time);
ans=query(rt[time],1,n,l,r);
printf("%lld\n",ans);
}
else
{
scanf("%d",&time);
t=time;
}
}
}
return 0;
}