链接：戳这里
To the moon Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) 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 1 Sample Output 4 55 9 15 0 1
题意：
长度为n的整数序列，支持四个操作
1：Q l r 输出区间[l,r]的总和
2：C l r x 区间[l,r]的每个值都增加x，此时时间增加1
3：H l r t 询问在t时刻区间[l,r]的总和
4：B t 时间回到t
思路：
可持久化线段树区间询问的裸体
这题和这个是一样的，唯一的不同是内存限制变小了
今天学到了一种函数式线段树成段更新时节约内存的办法 。。。
先考虑朴素的仅支持成段加减的线段树，我们可以用方式解决：
1.正常的懒惰标记，当访问到带有懒惰标记节点的子区间时将标记下传；
2.不用下传的懒惰标记，我们用一个标记来记录当前节点的整段区间被累加了多少，当询问的时候我们在从根节点走到目标结点的过程中不断累加所经过节点上的标记值。。。
基于这两种思想，就有了函数式线段树的两种实现方式，第一种在pushdown的过程中会产生大量的结点，而第二种没有pushdown的过程，不会新建过多的节点
代码：
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<string> #include<vector> #include <ctime> #include<queue> #include<set> #include<map> #include<list> #include<stack> #include<iomanip> #include<cmath> #include<bitset> #define mst(ss,b) memset((ss),(b),sizeof(ss)) ///#pragma comment(linker, "/STACK:102400000,102400000") typedef long long ll; typedef long double ld; #define INF (1ll<<60)-1 #define Max 30*100000 using namespace std; int n,m; ll a[100100]; struct node{ ll sum,lazy; int l,r; }tr[Max]; int tot=0; void build(int &root,int l,int r){ root=++tot; tr[root].l=l; tr[root].r=r; tr[root].lazy=0; if(l==r) { tr[root].sum=a[l]; return ; } int mid=(l+r)/2; build(tr[root].l,l,mid); build(tr[root].r,mid+1,r); tr[root].sum=tr[tr[root].l].sum+tr[tr[root].r].sum; } void update(int &root,int l,int r,int last,int x,int y,ll v){ root=++tot; tr[root]=tr[last]; tr[root].sum+=1LL*(y-x+1)*v; //printf("root=%d l=%d r=%d sum=%I64d x=%d y=%d\n",root,tr[root].l,tr[root].r,tr[root].sum,x,y); if(x==l && y==r){ tr[root].lazy+=v; return ; } int mid=(l+r)/2; if(y<=mid) update(tr[root].l,l,mid,tr[last].l,x,y,v); else if(x>mid) update(tr[root].r,mid+1,r,tr[last].r,x,y,v); else { update(tr[root].l,l,mid,tr[last].l,x,mid,v); update(tr[root].r,mid+1,r,tr[last].r,mid+1,y,v); } } ll query(int root,int l,int r,int x,int y){ ll ans=tr[root].lazy*1LL*(y-x+1); //printf("root=%d l=%d r=%d sum=%I64d x=%d y=%d\n",root,tr[root].l,tr[root].r,tr[root].sum,x,y); if(x==l && y==r){ return tr[root].sum; } int mid=(l+r)/2; if(y<=mid) ans+=query(tr[root].l,l,mid,x,y); else if(x>mid) ans+=query(tr[root].r,mid+1,r,x,y); else { ans+=query(tr[root].l,l,mid,x,mid); ans+=query(tr[root].r,mid+1,r,mid+1,y); } return ans; } int root[Max]; char s[110]; int main(){ while(scanf("%d%d",&n,&m)!=EOF){ tot=0; for(int i=1;i<=n;i++) scanf("%I64d",&a[i]); build(root[0],1,n); int now=0; for(int i=1;i<=m;i++){ getchar(); int x,y,t; ll v; scanf("%s",s); if(s[0]=='Q') { scanf("%d%d",&x,&y); printf("%I64d\n",query(root[now],1,n,x,y)); } else if(s[0]=='C'){ scanf("%d%d%I64d",&x,&y,&v); now++; update(root[now],1,n,root[now-1],x,y,v); } else if(s[0]=='H'){ scanf("%d%d%d",&x,&y,&t); printf("%I64d\n",query(root[t],1,n,x,y)); } else { scanf("%d",&t); now=t; } } } return 0; }
