[SPOJ-TTM]To the moon
题面
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.Input
n m
A 1 A 2 ... A n ... (here following the m operations. )Output
... (for each query, simply print the result. )
Sample Input
10 51 2 3 4 5 6 7 8 9 10Q 4 4Q 1 10Q 2 4C 3 6 3Q 2 4
思路
因为这道题不依赖前缀查询，所以我们可以像线段树一样在建树的时候就赋上值。
对于操作1,我们可以打个tag。注意，这里为了方便写一个永久tag，不下传，只在求和的时候直接累加。（具体看代码）
- 对于操作2,我们可以直接像线段树一样查询。
- 对于操作3,在时刻\(i\),我们只需要在2的基础上把初始节点改成\(rt[i]\)即可。
把记录最新版本的指针指向还原时刻\(t\)即可。
代码
#include#include #include #include #include #include #define ll long long#define maxn (int)(1e5+2)using namespace std;int n,m,rt[maxn],rs[maxn*29],ls[maxn*29];ll sum[maxn*29];int tag[maxn*29],idx;int build(int l,int r){ int now=++idx; if(l==r){scanf("%lld",&sum[now]);return now;} int mid=(l+r)>>1; ls[now]=build(l,mid);rs[now]=build(mid+1,r);sum[now]=sum[ls[now]]+sum[rs[now]]; return now;}int update(int pre,int l,int r,int tl,int tr,ll add){ int now=++idx; rs[now]=rs[pre],ls[now]=ls[pre],tag[now]=tag[pre]; sum[now]=sum[pre]+add*(min(r,tr)-max(tl,l)+1); int mid=(l+r)>>1; if(tl<=l&&r<=tr){tag[now]=tag[pre]+add;return now;} if(tl<=mid){ls[now]=update(ls[pre],l,mid,tl,tr,add);} if(tr>=mid+1){rs[now]=update(rs[pre],mid+1,r,tl,tr,add);} return now;}ll query(int now,int l,int r,int tl,int tr,ll add){ if(tl<=l&&r<=tr){return add*(r-l+1)+sum[now];} add+=tag[now];ll ans=0;int mid=(l+r)>>1; if(tl<=mid)ans+=query(ls[now],l,mid,tl,tr,add); if(tr>=mid+1)ans+=query(rs[now],mid+1,r,tl,tr,add); return ans;}int main(){ //freopen("in","r",stdin); int t=0;int flag=1; while(~scanf("%d%d",&n,&m)){ t=0,idx=0; if(!flag)printf("\n");flag=0; memset(tag,0,sizeof(tag)); memset(sum,0,sizeof(sum)); rt[0]=build(1,n); for(int i=1;i<=m;i++){ char s[3];int l,r;ll d; scanf("%s",s); if(s[0]=='C'){ scanf("%d%d%lld",&l,&r,&d); t++;rt[t]=update(rt[t-1],1,n,l,r,d); } else if(s[0]=='Q'){ scanf("%d%d",&l,&r); printf("%lld\n",query(rt[t],1,n,l,r,0)); } else if(s[0]=='H'){ scanf("%d%d%lld",&l,&r,&d); printf("%lld\n",query(rt[d],1,n,l,r,0)); } else if(s[0]=='B'){ scanf("%lld",&d);t=d; } } } return 0;}