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  • [SPOJ-TTM]To the moon

    [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 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
    

    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<string>
    #include<cstring>
    #include<iostream>
    #include<cstdio>
    #include<algorithm>
    #include<cmath>
    #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("
    ");
    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
    ",query(rt[t],1,n,l,r,0));
            }
            else if(s[0]=='H'){
                scanf("%d%d%lld",&l,&r,&d);
                printf("%lld
    ",query(rt[d],1,n,l,r,0));
            }
            else if(s[0]=='B'){
                scanf("%lld",&d);t=d;
            }
        }
       
        }
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/GavinZheng/p/10840559.html
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