To the moon
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 5117 Accepted Submission(s): 1152
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.
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. )
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
Author
HIT
Source
题意:对于一个长度为n的序列 执行4种操作
C l r d 区间[l,r]的数全部增加d 并且当前时刻增加一
Q l r 求区间[l,r]的和
H l r t 求第t个时刻
B t 返回第t个时刻
题解:据说是主席树区间更新入门题目。
1 #pragma comment(linker, "/STACK:1024000000,1024000000") 2 #include <iostream> 3 #include <cstdio> 4 #include <cstdlib> 5 #include <cstring> 6 #include <algorithm> 7 #include <stack> 8 #include <queue> 9 #include <cmath> 10 #include <map> 11 #define ll __int64 12 #define mod 1000000007 13 #define dazhi 2147483647 14 #define N 530005 15 using namespace std; 16 ll a[N]; 17 char s[10]; 18 struct chairmantree 19 { 20 int rt[20*N],ls[20*N],rs[20*N]; 21 ll sum[N*20],lazy[20*N]; 22 int tot; 23 void init() 24 { 25 tot=0; 26 } 27 void pushup(int l,int r,int pos) 28 { 29 sum[pos]=sum[ls[pos]]+sum[rs[pos]]+1LL*(r-l+1)*lazy[pos]; 30 } 31 void buildtree(int l,int r,int &pos) 32 { 33 pos=++tot; 34 lazy[pos]=0; 35 sum[pos]=0; 36 if(l==r) 37 { 38 sum[pos]=a[l]; 39 return ; 40 } 41 int mid=(l+r)>>1; 42 buildtree(l,mid,ls[pos]); 43 buildtree(mid+1,r,rs[pos]); 44 pushup(l,r,pos); 45 } 46 void update(int L,int R,ll c,int pre,int l,int r,int &pos) 47 { 48 pos=++tot; 49 ls[pos]=ls[pre]; 50 rs[pos]=rs[pre]; 51 sum[pos]=sum[pre]; 52 lazy[pos]=lazy[pre]; 53 if(L==l&&R==r) 54 { 55 sum[pos]+=1LL*(r-l+1)*c; 56 lazy[pos]+=c; 57 return ; 58 } 59 int mid=(l+r)>>1; 60 if(R<=mid) 61 update(L,R,c,ls[pre],l,mid,ls[pos]); 62 else 63 { 64 if(L>mid) 65 update(L,R,c,rs[pre],mid+1,r,rs[pos]); 66 else 67 { 68 update(L,mid,c,ls[pre],l,mid,ls[pos]); 69 update(mid+1,R,c,rs[pre],mid+1,r,rs[pos]); 70 } 71 } 72 pushup(l,r,pos); 73 } 74 ll query(int L,int R,int l,int r,int pos) 75 { 76 if(L==l&&R==r) 77 return sum[pos]; 78 int mid=(l+r)>>1; 79 ll ans=1LL*lazy[pos]*(R-L+1); 80 if(R<=mid) 81 ans+=query(L,R,l,mid,ls[pos]); 82 else 83 { 84 if(L>mid) 85 ans+=query(L,R,mid+1,r,rs[pos]); 86 else 87 { 88 ans+=query(L,mid,l,mid,ls[pos]); 89 ans+=query(mid+1,R,mid+1,r,rs[pos]); 90 } 91 } 92 return ans; 93 } 94 } tree; 95 int main() 96 { 97 int n,m; 98 while(scanf("%d %d",&n,&m)!=EOF) 99 { 100 for(int i=1; i<=n; i++) 101 scanf("%I64d",&a[i]); 102 tree.init(); 103 tree.buildtree(1,n,tree.rt[0]); 104 int now=0; 105 while(m--) 106 { 107 scanf("%s",s); 108 if(s[0]=='C') 109 { 110 int l,r; 111 ll c; 112 scanf("%d%d%I64d",&l,&r,&c); 113 tree.update(l,r,c,tree.rt[now],1,n,tree.rt[now+1]); 114 now++; 115 } 116 if(s[0]=='Q') 117 { 118 int l,r; 119 scanf("%d %d",&l,&r); 120 printf("%I64d ",tree.query(l,r,1,n,tree.rt[now])); 121 } 122 if(s[0]=='H') 123 { 124 int l,r,t; 125 scanf("%d %d %d",&l,&r,&t); 126 printf("%I64d ",tree.query(l,r,1,n,tree.rt[t])); 127 } 128 if(s[0]=='B') 129 { 130 int x; 131 scanf("%d",&x); 132 now=x; 133 } 134 } 135 } 136 return 0; 137 }