HDU 4578 - Transformation - [加强版线段树]
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4578
Problem Description
Yuanfang is puzzled with the question below:
There are n integers, a1, a2, …, an. The initial values of them are 0. There are four kinds of operations.
Operation 1: Add c to each number between ax and ay inclusive. In other words, do transformation ak<---ak+c, k = x,x+1,…,y.
Operation 2: Multiply c to each number between ax and ay inclusive. In other words, do transformation ak<---ak×c, k = x,x+1,…,y.
Operation 3: Change the numbers between ax and ay to c, inclusive. In other words, do transformation ak<---c, k = x,x+1,…,y.
Operation 4: Get the sum of p power among the numbers between ax and ay inclusive. In other words, get the result of axp+ax+1p+…+ay p.
Yuanfang has no idea of how to do it. So he wants to ask you to help him.
There are n integers, a1, a2, …, an. The initial values of them are 0. There are four kinds of operations.
Operation 1: Add c to each number between ax and ay inclusive. In other words, do transformation ak<---ak+c, k = x,x+1,…,y.
Operation 2: Multiply c to each number between ax and ay inclusive. In other words, do transformation ak<---ak×c, k = x,x+1,…,y.
Operation 3: Change the numbers between ax and ay to c, inclusive. In other words, do transformation ak<---c, k = x,x+1,…,y.
Operation 4: Get the sum of p power among the numbers between ax and ay inclusive. In other words, get the result of axp+ax+1p+…+ay p.
Yuanfang has no idea of how to do it. So he wants to ask you to help him.
Input
There are no more than 10 test cases.
For each case, the first line contains two numbers n and m, meaning that there are n integers and m operations. 1 <= n, m <= 100,000.
Each the following m lines contains an operation. Operation 1 to 3 is in this format: "1 x y c" or "2 x y c" or "3 x y c". Operation 4 is in this format: "4 x y p". (1 <= x <= y <= n, 1 <= c <= 10,000, 1 <= p <= 3)
The input ends with 0 0.
For each case, the first line contains two numbers n and m, meaning that there are n integers and m operations. 1 <= n, m <= 100,000.
Each the following m lines contains an operation. Operation 1 to 3 is in this format: "1 x y c" or "2 x y c" or "3 x y c". Operation 4 is in this format: "4 x y p". (1 <= x <= y <= n, 1 <= c <= 10,000, 1 <= p <= 3)
The input ends with 0 0.
Output
For each operation 4, output a single integer in one line representing the result. The answer may be quite large. You just need to calculate the remainder of the answer when divided by 10007.
Sample Input
5 5
3 3 5 7
1 2 4 4
4 1 5 2
2 2 5 8
4 3 5 3
0 0
Sample Output
307
7489
题意：
给出一个序列，有下列操作：
- 对区间[x,y]全部加上c
- 对区间[x,y]全部乘上c
- 将区间[x,y]全部改成c
- 查询区间[x,y]的p次方和
题解：
加强版的线段树，需要三个lazy标记，一个add表示加法标记，一个mul表示乘法标记，一个alt表示修改标记，
同时由于p=1,2,3，所以可以有三个val值：sum1表示一次方和，sum2表示平方和，sum3表示立方和，
然后我们要确定三个标记的优先级：alt第一，mul第二，add第三，pushdown的时候要按照这样的顺序pushdown，
同时下压高优先级的标记，会影响到低优先级的标记，这个需要注意，
另外，在接收到父节点传过来的add标记时，更新自身时（update_add成员函数），要注意计算sum3，sum2，sum1的先后顺序，一定是sum3，sum2，sum1，
这三个sum计算的方法如下：
$\begin{array}{l} \left( {a + x} \right)^2 = a^2 + 2ax + x^2 \\ \left( {a_1 + x} \right)^2 + \left( {a_2 + x} \right)^2 + \cdots + \left( {a_n + x} \right)^2 = \left( {a_1 ^2 + \cdots + a_n ^2 } \right) + 2x\left( {a_1 + \cdots + a_n } \right) + nx^2 \\ \left( {a + x} \right)^3 = a^3 + 3a^2 x + 3ax^2 + x^3 \\ \left( {a_1 + x} \right)^3 + \left( {a_2 + x} \right)^3 + \cdots + \left( {a_n + x} \right)^3 = \left( {a_1 ^3 + \cdots + a_n ^3 } \right) + 3x\left( {a_1 ^2 + \cdots + a_n ^2 } \right) + 3x^2 \left( {a_1 + \cdots + a_n } \right) + nx^3 \\ \end{array}$
AC代码：
#include<bits/stdc++.h> using namespace std; typedef long long ll; const int maxn=100000+10; const int MOD=10007; int n,m; /********************************* Segment Tree - st *********************************/ struct Node { int l,r; int sum1,sum2,sum3; int add,mul,alt; void Update_Alt(int x) { x%=MOD; sum1 = (r-l+1) * x % MOD; sum2 = (r-l+1) * x % MOD * x % MOD; sum3 = (r-l+1) * x % MOD * x % MOD * x % MOD; alt=x; add=0; mul=1; } void Update_Mul(int x) { x%=MOD; sum1 = sum1 % MOD * x % MOD; sum2 = sum2 % MOD * x % MOD * x % MOD; sum3 = sum3 % MOD * x % MOD * x % MOD * x % MOD; mul = mul % MOD * x % MOD; add = add % MOD * x % MOD; } void Update_Add(int x) { x%=MOD; sum3 = ( sum3%MOD + 3*x%MOD*sum2%MOD + 3*x%MOD*x%MOD*sum1%MOD + (r-l+1)*x%MOD*x%MOD*x%MOD ) % MOD; sum2 = ( sum2%MOD + 2*x%MOD*sum1%MOD + (r-l+1)%MOD*x%MOD*x%MOD ) % MOD; sum1 = ( sum1%MOD + (r-l+1)%MOD*x%MOD ) % MOD; add=(add%MOD+x)%MOD; } }node[4*maxn]; void Pushdown(int root) { int ls=root*2, rs=root*2+1; if(node[root].alt!=0) { node[ls].Update_Alt(node[root].alt); node[rs].Update_Alt(node[root].alt); node[root].alt=0; } if(node[root].mul!=1) { node[ls].Update_Mul(node[root].mul); node[rs].Update_Mul(node[root].mul); node[root].mul=1; } if(node[root].add!=0) { node[ls].Update_Add(node[root].add); node[rs].Update_Add(node[root].add); node[root].add=0; } } void Pushup(int root) { int ls=root*2, rs=root*2+1; node[root].sum1=(node[ls].sum1+node[rs].sum1)%MOD; node[root].sum2=(node[ls].sum2+node[rs].sum2)%MOD; node[root].sum3=(node[ls].sum3+node[rs].sum3)%MOD; } void Build(int root,int l,int r) //对区间[l,r]建树 { if(l>r) return; node[root].l=l; node[root].r=r; node[root].sum1=0; node[root].sum2=0; node[root].sum3=0; node[root].alt=0; node[root].add=0; node[root].mul=1; if(l<r) { int mid=l+(r-l)/2; Build(root*2,l,mid); Build(root*2+1,mid+1,r); Pushup(root); } } void Alt(int root,int st,int ed,ll val) //区间[st,ed]全部改成val { if(st>node[root].r || ed<node[root].l) return; if(st<=node[root].l && node[root].r<=ed) node[root].Update_Alt(val); else { Pushdown(root); Alt(root*2,st,ed,val); Alt(root*2+1,st,ed,val); Pushup(root); } } void Mul(int root,int st,int ed,ll val) //区间[st,ed]全部加上val { if(st>node[root].r || ed<node[root].l) return; if(st<=node[root].l && node[root].r<=ed) node[root].Update_Mul(val); else { Pushdown(root); Mul(root*2,st,ed,val); Mul(root*2+1,st,ed,val); Pushup(root); } } void Add(int root,int st,int ed,ll val) //区间[st,ed]全部加上val { if(st>node[root].r || ed<node[root].l) return; if(st<=node[root].l && node[root].r<=ed) node[root].Update_Add(val); else { Pushdown(root); Add(root*2,st,ed,val); Add(root*2+1,st,ed,val); Pushup(root); } } int Query(int root,int st,int ed,int p) //查询区间[st,ed]的p次方和 { if(st>node[root].r || ed<node[root].l) return 0; if(st<=node[root].l && node[root].r<=ed) { if(p==1) return node[root].sum1; if(p==2) return node[root].sum2; if(p==3) return node[root].sum3; } else { Pushdown(root); int ls=Query(root*2,st,ed,p)%MOD; int rs=Query(root*2+1,st,ed,p)%MOD; Pushup(root); return (ls+rs)%MOD; } } /********************************* Segment Tree - st *********************************/ int main() { while(scanf("%d%d",&n,&m) && n*m!=0) { Build(1,1,n); for(int i=1;i<=m;i++) { int op; scanf("%d",&op); if(op==1) { int x,y,k; scanf("%d%d%d",&x,&y,&k); Add(1,x,y,k); } if(op==2) { int x,y,k; scanf("%d%d%d",&x,&y,&k); Mul(1,x,y,k); } if(op==3) { int x,y,k; scanf("%d%d%d",&x,&y,&k); Alt(1,x,y,k); } if(op==4) { int l,r,p; scanf("%d%d%d",&l,&r,&p); printf("%d\n",Query(1,l,r,p)); } } } }
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