Poj 3468-A Simple Problem with Integers 线段树,树状数组
A Simple Problem with Integers
| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 85851 | Accepted: 26685 | |
| Case Time Limit: 2000MS | ||
Description
You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval.
Input
The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000.
The second line contains N numbers, the initial values of A1, A2, ... , AN. -1000000000 ≤ Ai ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of Aa, Aa+1, ... , Ab. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of Aa, Aa+1, ... , Ab.
Output
You need to answer all Q commands in order. One answer in a line.
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
Sample Output
4 55 9 15
Hint
The sums may exceed the range of 32-bit integers.
Source
POJ Monthly--2007.11.25, Yang Yi
题意:给一串数字，每次区间加上一个数，或询问区间和。
题解:
线段树水过，打个lazy标记就好。。。
线段树程序：
1 #include<cstdio> 2 #include<cstring> 3 #include<cstdlib> 4 #include<cmath> 5 #include<iostream> 6 #include<algorithm> 7 using namespace std; 8 #define LL long long 9 #define MAXN 100010 10 struct node 11 { 12 LL left,right,sum,tag; 13 }tree[MAXN*4]; 14 LL A[MAXN]; 15 LL read() 16 { 17 LL s=0,fh=1;char ch=getchar(); 18 while(ch<'0'||ch>'9'){if(ch=='-')fh=-1;ch=getchar();} 19 while(ch>='0'&&ch<='9'){s=s*10+(ch-'0');ch=getchar();} 20 return s*fh; 21 } 22 void Pushup(LL k) 23 { 24 tree[k].sum=tree[k*2].sum+tree[k*2+1].sum; 25 } 26 void Pushdown(LL k,LL l,LL r) 27 { 28 if(tree[k].tag!=0) 29 { 30 tree[k*2].tag+=tree[k].tag; 31 tree[k*2+1].tag+=tree[k].tag; 32 LL mid=(l+r)/2; 33 tree[k*2].sum+=tree[k].tag*(mid-l+1); 34 tree[k*2+1].sum+=tree[k].tag*(r-mid); 35 tree[k].tag=0; 36 } 37 } 38 void Build(LL k,LL l,LL r) 39 { 40 tree[k].left=l;tree[k].right=r; 41 if(l==r) 42 { 43 tree[k].sum=A[l]; 44 return; 45 } 46 LL mid=(l+r)/2; 47 Build(k*2,l,mid);Build(k*2+1,mid+1,r); 48 Pushup(k); 49 } 50 void Add(LL k,LL ql,LL qr,LL add) 51 { 52 //Pushdown(k,tree[k].left,tree[k].right); 53 if(ql<=tree[k].left&&tree[k].right<=qr){tree[k].tag+=add;tree[k].sum+=(tree[k].right-tree[k].left+1)*add;return;} 54 Pushdown(k,tree[k].left,tree[k].right); 55 LL mid=(tree[k].left+tree[k].right)/2; 56 if(qr<=mid)Add(k*2,ql,qr,add); 57 else if(ql>mid)Add(k*2+1,ql,qr,add); 58 else {Add(k*2,ql,mid,add);Add(k*2+1,mid+1,qr,add);} 59 Pushup(k); 60 } 61 LL Sum(LL k,LL ql,LL qr) 62 { 63 //Pushdown(k,tree[k].left,tree[k].right); 64 if(ql<=tree[k].left&&tree[k].right<=qr)return tree[k].sum; 65 Pushdown(k,tree[k].left,tree[k].right); 66 LL mid=(tree[k].left+tree[k].right)/2; 67 if(qr<=mid)return Sum(k*2,ql,qr); 68 else if(ql>mid)return Sum(k*2+1,ql,qr); 69 else return Sum(k*2,ql,mid)+Sum(k*2+1,mid+1,qr); 70 } 71 int main() 72 { 73 LL N,Q,i,a,b,c; 74 char fh[2]; 75 N=read();Q=read(); 76 for(i=1;i<=N;i++)A[i]=read(); 77 Build(1,1,N); 78 for(i=1;i<=Q;i++) 79 { 80 scanf("\n%s",fh); 81 if(fh[0]=='Q') 82 { 83 a=read();b=read(); 84 printf("%lld\n",Sum(1,a,b)); 85 } 86 else 87 { 88 a=read();b=read();c=read(); 89 Add(1,a,b,c); 90 } 91 } 92 return 0; 93 }
树状数组不会。。。
先附上fhq神犇的题解：
上次NOI集训的时候，一位福建女神牛和我探讨过这题能不能用BIT，我当时
的答复是可以，因为“扩展树状数组”这个东西理论上可以实现一般线段树
可以实现的东西，且空间上的常数好一点。但是对于“扩展树状数组”，这
个东西是我一时兴起想到的玩意，没有进行更多的研究，没查到任何的资料
，更没有想过如何把线段树著名的lazy思想照搬上去，以及动态开内存的解
决方案。
权衡利弊，我想了一个使用两棵BIT的替代方法来解决这题，并且成功地将
内存使用做到了1728K。这恐怕是带标记的扩展树状数组达不到的。
记录两个BIT，
设数列为A[i],BIT1的每个元素B1[i]=A[i]*i,
BIT2的每个元素B2[i]=A[i]
则：sum{A[i]|i<=a}=sum{B1[i]|i<=a}+(sum{B2[i]|1<=i<=N}-sum{B2[i]|i<=a})*a
sum{A[i]|a<=i<=b}=sum{A[i]|i<=b}-sum{A[i]|i<a}
这样就十分巧妙的解决了！
关键代码：
int N;
struct BIT{
long long a[NMax];
void ins(int x,long long k){
for (;x<N;x+=((x+1)&-(x+1)))a[x]+=k;
}
long long ask(int x){
long long ret;
for (ret=0;x>=0;x-=((x+1)&-(x+1)))ret+=a[x];
return ret;
}
}B1,B2;
long long B2S;
void Add(int a,long long x){
//printf("Add %d %I64d\n",a,x);
B1.ins(a,x*((long long)a+1));
B2.ins(a,x);B2S+=x;
}
void Add(int a,int b,long long x){
Add(b,x);
if (a)Add(a-1,-x);
}
long long Ask(int a){
//printf("Ask %d\n",a);
long long ret=B1.ask(a)+(B2S-B2.ask(a))*(long long)(a+1);
//printf("=%I64d\n",ret);
return ret;
}
long long Ask(int a,int b){
long long ret;
ret=Ask(b);
if (a)ret-=Ask(a-1);
return ret;
}
填坑中。。。。。。
