| Time Limit: 2000MS | Memory Limit: 65536KB | 64bit IO Format: %I64d & %I64u |
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 {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.
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 {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.
Input
n m
A 1 A 2 ... A n
... (here following the m operations. )
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
Source
可持久化线段树模板题。
对于不同的时间建立不同的新结点，新结点按照线段树的规则连接各个被修改的结点的新址（修改时不在原结点修改，而是新建一个结点（类似于分层图什么的）），没有修改的区域就直接链接到旧树的结点（因此不能用root*2 root*2+1的方式算结点，而要用数组模拟指针记录子结点标号）。
由于新层和旧层修改的值不一样，所以lazy标记是不能持久化的，一层的lazy只能在一层用。具体的解决方法见代码。
↑看到有大神说lazy标记持久化的方法是，对每个lazy记录时间戳，只有其标记时间与当前所要求的状态的时间相同时，才计算。然而看上去好麻烦。
代码基本靠抄。
看到有人说这题卡内存，就特意把数组开小了，结果依旧MLE。折腾一个多小时无果，怒把数组开到300w，居然A了。
↑想了想，大概MLE是因为动态申请新节点时，因为t数组越界，申请到了超大的节点值，直接炸掉内存，所以MLE而不是RE。
感到内存学问博大精深。
1 /*by SilverN*/ 2 #include<algorithm> 3 #include<iostream> 4 #include<cstring> 5 #include<cstdio> 6 #include<cmath> 7 #include<vector> 8 #define LL long long 9 #define mid (l+r)/2 10 using namespace std; 11 const int mxn=100010; 12 int read(){ 13 int x=0,f=1;char ch=getchar(); 14 while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();} 15 while(ch>='0' && ch<='9'){x=x*10+ch-'0';ch=getchar();} 16 return x*f; 17 } 18 int n,m; 19 int data[mxn]; 20 struct node{ 21 int lc,rc; 22 LL sum,mk; 23 }t[3000010]; 24 int root[mxn]; 25 int nct=0; 26 int ntime=0; 27 void Build(int l,int r,int &rt){ 28 t[++nct]=t[rt]; 29 rt=nct; 30 if(l==r){ 31 t[rt].sum=data[l]; 32 return; 33 } 34 Build(l,mid,t[rt].lc); 35 Build(mid+1,r,t[rt].rc); 36 t[rt].sum=t[t[rt].lc].sum+t[t[rt].rc].sum; 37 return; 38 } 39 void add(int L,int R,int v,int l,int r,int &rt){ 40 t[++nct]=t[rt]; 41 rt=nct; 42 t[rt].sum+=(LL)v*(min(R,r)-max(l,L)+1); 43 if(L<=l && r<=R){ 44 t[rt].mk+=v; 45 return; 46 } 47 if(L<=mid)add(L,R,v,l,mid,t[rt].lc); 48 if(R>mid)add(L,R,v,mid+1,r,t[rt].rc); 49 return; 50 } 51 LL query(int L,int R,int l,int r,int rt){ 52 if(L<=l && r<=R)return t[rt].sum; 53 LL res=t[rt].mk*1LL*(min(R,r)-(max(L,l))+1); 54 if(L<=mid)res+=query(L,R,l,mid,t[rt].lc); 55 if(R>mid)res+=query(L,R,mid+1,r,t[rt].rc); 56 return res; 57 } 58 int main(){ 59 char op[3]; 60 int i,j,x,y,a; 61 while(~scanf("%d%d",&n,&m)){ 62 ntime=0;nct=0; 63 for(i=1;i<=n;i++) 64 data[i]=read(); 65 Build(1,n,root[0]); 66 for(i=1;i<=m;i++){ 67 scanf("%s",op); 68 switch(op[0]){ 69 case 'Q':{ 70 x=read();y=read(); 71 printf("%lld ",query(x,y,1,n,root[ntime])); 72 break; 73 } 74 case 'C':{ 75 x=read();y=read();a=read(); 76 ++ntime; 77 root[ntime]=root[ntime-1]; 78 add(x,y,a,1,n,root[ntime]); 79 break; 80 } 81 case 'H':{ 82 x=read();y=read();a=read(); 83 printf("%lld ",query(x,y,1,n,root[a])); 84 break; 85 } 86 case 'B':{ 87 ntime=read(); 88 break; 89 } 90 } 91 } 92 } 93 return 0; 94 }