HDU 4348 / SPOJ TTM To the moon [主席树]

Description

You‘ve been given N integers A[1], A[2],..., A[N]. On these integers, you need to implement the following operations:

• C l r d: Adding a constant d for every {Ai | l <= i <= r}, and increase the timestamp by 1, this is the only operation that will cause the timestamp increase. 
• Q l r: Querying the current sum of {Ai | l <= i <= r}.
• H l r t: Querying a history sum of {Ai | l <= i <= r} in time t.
• 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. 

题意：给出N个数的序列，有四种操作：

1.成段区间加D，此时的序列有一个时间T来表示它（每次操作1之后T+1）

2.询问当前序列区间和

3.询问T时刻序列的区间和

4.回到T时刻（这里比较坑，题意不是很清楚，下次T+1是从回到的时间开始，WA了无数遍）

解法：主席树可写，时空复杂度为NlogNlogN，算是模板题了，不赘述。考虑了一下，其实还可以线段树离线处理询问直接搞，时间复杂度为NlogN，空间为N，将T时刻的询问先存起来，等到了T时刻的时候直接把相关询问处理完(需要区分是第几次到T时刻)，返回T时刻的时候，将前面做过的区间加重新减回去即可，因为每个相反的操作最多重新做一遍，所以复杂度依旧是NlogN。

主席树代码：

#include<stdio.h>#include<string.h>#include<algorithm>#include<math.h>#include<iostream>#include<stdlib.h>#include<set>#include<map>#include<queue>#include<vector>#include<bitset>#pragma comment(linker, "/STACK:1024000000,1024000000")template <class T>bool scanff(T &ret){ //Faster Input    char c; int sgn; T bit=0.1;    if(c=getchar(),c==EOF) return 0;    while(c!='-'&&c!='.'&&(c<'0'||c>'9')) c=getchar();    sgn=(c=='-')?-1:1;    ret=(c=='-')?0:(c-'0');    while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');    if(c==' '||c=='\n'){ ret*=sgn; return 1; }    while(c=getchar(),c>='0'&&c<='9') ret+=(c-'0')*bit,bit/=10;    ret*=sgn;    return 1;}#define inf 1073741823#define llinf 4611686018427387903LL#define PI acos(-1.0)#define lth (th<<1)#define rth (th<<1|1)#define rep(i,a,b) for(int i=int(a);i<=int(b);i++)#define drep(i,a,b) for(int i=int(a);i>=int(b);i--)#define gson(i,root) for(int i=ptx[root];~i;i=ed[i].next)#define tdata int testnum;scanff(testnum);for(int cas=1;cas<=testnum;cas++)#define mem(x,val) memset(x,val,sizeof(x))#define mkp(a,b) make_pair(a,b)#define findx(x) lower_bound(b+1,b+1+bn,x)-b#define pb(x) push_back(x)using namespace std;typedef long long ll;typedef pair<int,int> pii;#define NN 100100#define MM 2500000int n,qn,ql,qr;int tot;int lson[MM],rson[MM];ll c[MM],add[MM];ll a[NN];int t[NN];int build(int l,int r){    int x=++tot;    if(l!=r){        int m=(l+r)>>1;        lson[x]=build(l,m);        rson[x]=build(m+1,r);        c[x]=c[lson[x]]+c[rson[x]];    }    else c[x]=a[l];    return x;}void update(int l,int r,int pre,int cur,ll val){    lson[cur]=lson[pre];    rson[cur]=rson[pre];    c[cur]=c[pre];    add[cur]=add[pre];    if(l>=ql&&r<=qr){        add[cur]+=val;        if(l==r)c[cur]=a[l]+add[cur]*(r-l+1);        else c[cur]=c[lson[cur]]+c[rson[cur]]+add[cur]*(r-l+1);        return ;    }    int m=(l+r)>>1;    if(m>=ql)update(l,m,lson[pre],lson[cur]=++tot,val);    if(m<qr)update(m+1,r,rson[pre],rson[cur]=++tot,val);    c[cur]=c[lson[cur]]+c[rson[cur]]+add[cur]*(r-l+1);}ll query(int l,int r,int x,ll d){    if(l>=ql&&r<=qr)return c[x]+(r-l+1)*d;    int m=(l+r)>>1;    ll sum=0;    if(m>=ql)sum+=query(l,m,lson[x],d+add[x]);    if(m<qr)sum+=query(m+1,r,rson[x],d+add[x]);    return sum;}char op[111];int main(){    scanff(n);    scanff(qn);    rep(i,1,n)scanff(a[i]);    build(1,n);    t[0]=1;    int pre=0;    int cur=0;    rep(i,1,qn){        scanf("%s",op);        if(op[0]=='C'){            ll val;            scanff(ql);            scanff(qr);            scanff(val);            t[++cur]=++tot;            update(1,n,t[pre],t[cur],val);            pre=cur;        }        if(op[0]=='B'){            scanff(pre);            cur=pre;        }        if(op[0]=='Q'){            scanff(ql);            scanff(qr);            printf("%lld\n",query(1,n,t[pre],0));        }        if(op[0]=='H'){            int k;            scanff(ql);            scanff(qr);            scanff(k);            printf("%lld\n",query(1,n,t[k],0));        }    }}/*10 10001 2 3 4 5 6 7 8 9 10Q 4 4Q 1 10Q 2 4C 3 6 3Q 2 4*/