SPOJ 11840. Sum of Squares with Segment Tree (线段树，区间更新)

http://www.spoj.com/problems/SEGSQRSS/

SPOJ Problem Set (classical)

11840. Sum of Squares with Segment Tree

Problem code: SEGSQRSS

Segment trees are extremely useful.  In particular "Lazy Propagation" (i.e. see here, for example) allows one to compute sums over a range in O(lg(n)), and update ranges in O(lg(n)) as well.  In this problem you will compute something much harder:  

The sum of squares over a range with range updates of 2 types:

1) increment in a range

2) set all numbers the same in a range.

Input

There will be T (T <= 25) test cases in the input file.  First line of the input contains two positive integers, N (N <= 100,000) and Q (Q <= 100,000). The next line contains N integers, each at most 1000.  Each of the next Qlines starts with a number, which indicates the type of operation:

2 st nd  -- return the sum of the squares of the numbers with indices in [st, nd] {i.e., from st to nd inclusive} (1 <= st <= nd <= N).

1 st nd x -- add "x" to all numbers with indices in [st, nd] (1 <= st <= nd <= N, and -1,000 <= x <= 1,000).

0 st nd x -- set all numbers with indices in [st, nd] to "x" (1 <= st <= nd <= N, and -1,000 <= x <= 1,000).

 

Output

For each test case output the “Case <caseno>:” in the first line and from the second line output the sum of squares for each operation of type 2.  Intermediate overflow will not occur with proper use of 64-bit signed integer. 

Example

Input:

2
4 5
1 2 3 4
2 1 4
0 3 4 1
2 1 4
1 3 4 1
2 1 4
1 1
1
2 1 1

Output:

Case 1:
30
7
13
Case 2:
1

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Added by:Chen XiaohongDate:2012-07-11Time limit:6sSource limit:50000BMemory limit:256MBCluster:Pyramid (Intel Pentium III 733 MHz)Languages:All

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题意：

有三种操作：将区间中的所有数置为x；将区间中的所有数加上x；求区间内所有数的平方和。

分析：

先考虑如果不需要求平方和，只是求和，我们需要维护这些数据：addv-区间内的数共同加上的值;setv-区间内的数都置为的值(setv=INF表示不设置);sumv-区间内的数加上addv之前的值。

但这题求的是平方和，似乎不是很好维护。如果只是set操作，还是很好维护的，那么难点就在于add操作了。考虑如下等式：(x+v)^2=x^2+2xv+v^2,x是add操作之前的数,v是add的数，这是一个数的情况，那么一段区间内的数呢？

显然有sum(xi^2)+(v^2)*length+2*sum(xi)*v，这样问题就迎刃而解了，只要再维护一个区间的平方和就行，当set时直接改，add时加上(v^2)*length+2*sum(xi)*v即可。

/* * * Author : fcbruce * * Time : Fri 03 Oct 2014 04:16:10 PM CST * */#include <cstdio>#include <iostream>#include <sstream>#include <cstdlib>#include <algorithm>#include <ctime>#include <cctype>#include <cmath>#include <string>#include <cstring>#include <stack>#include <queue>#include <list>#include <vector>#include <map>#include <set>#define sqr(x) ((x)*(x))#define LL long long#define itn int#define INF 0x3f3f3f3f#define PI 3.1415926535897932384626#define eps 1e-10#ifdef _WIN32  #define lld "%I64d"#else  #define lld "%lld"#endif#define maxm #define maxn 100007using namespace std;int addv[maxn<<2],setv[maxn<<2];long long sumv[maxn<<2],sqrsumv[maxn<<2];inline void pushdown(int k,int l,int r){  int lc=k*2+1,rc=k*2+2,m=l+r>>1;  addv[lc]+=addv[k];  addv[rc]+=addv[k];  addv[k]=0;  if (setv[k]!=INF)  {    setv[lc]=setv[rc]=setv[k];    sumv[lc]=(LL)setv[lc]*(m-l);sumv[rc]=(LL)setv[rc]*(r-m);    sqrsumv[lc]=sqr((LL)setv[k]*(m-l));sqrsumv[rc]=sqr((LL)setv[rc])*(r-m);    addv[lc]=addv[rc]=0;    setv[k]=INF;  }}inline void pushup(int k,int l,int r){  int lc=k*2+1,rc=k*2+2,m=l+r>>1;  sumv[k]=sumv[lc]+sumv[rc]+(LL)addv[lc]*(m-l)+(LL)addv[rc]*(r-m);  sqrsumv[k]=sqrsumv[lc]+sqrsumv[rc]+(LL)(r-l)*(addv[k])+2ll*sumv[k]*addv[k];}void build(int k,int l,int r){  addv[k]=0;  setv[k]=INF;  sumv[k]=sqrsumv[k]=0ll;  if (r-l==1)  {    scanf("%d",&sumv[k]);    sqrsumv[k]=sqr((LL)sumv[k]);    return ;  }  build(k*2+1,l,l+r>>1);  build(k*2+2,l+r>>1,r);  pushup(k,l,r);}void update_add(int a,int b,int v,int k,int l,int r){  if (b<=l || r<=a) return ;  if (a<=l && r<=b)  {    addv[k]+=v;    sqrsumv[k]+=sqr((LL)v)*(r-l)+2ll*v*sumv[k];    return ;  }  pushdown(k,l,r);  update_add(a,b,v,k*2+1,l,l+r>>1);  update_add(a,b,v,k*2+2,l+r>>1,r);  pushup(k,l,r);}void update_set(int a,int b,int v,int k,int l,int r){  if (b<=l || r<=a) return ;  if (a<=l && r<=b)  {    addv[k]=0;    setv[k]=v;    sumv[k]=(LL)v*(r-l);    sqrsumv[k]=sqr((LL)v)*(r-l);    return ;  }  pushdown(k,l,r);  update_set(a,b,v,k*2+1,l,l+r>>1);  update_set(a,b,v,k*2+2,l+r>>1,r);  pushup(k,l,r);}long long query(int a,int b,int k,int l,int r){  if (b<=l || r<=a) return 0ll;  if (a<=l && r<=b) return sqrsumv[k];    pushdown(k,l,r);  return query(a,b,k*2+1,l,l+r>>1)+query(a,b,k*2+2,l+r>>1,r);}int main(){#ifdef FCBRUCE  freopen("/home/fcbruce/code/t","r",stdin);#endif // FCBRUCE  int T_T,__=0;  scanf("%d",&T_T);  while (T_T--)  {    int n,m;    scanf("%d%d",&n,&m);        build(0,0,n);    printf("Case %d:\n",++__);    int op,a,b,v;    while (m--)    {      scanf("%d",&op);      switch (op)      {        case 0:          scanf("%d%d%d",&a,&b,&v);          a--;          update_set(a,b,v,0,0,n);          break;        case 1:          scanf("%d%d%d",&a,&b,&v);          a--;          update_add(a,b,v,0,0,n);          break;        case 2:          scanf("%d %d",&a,&b);          a--;          printf(lld "\n",query(a,b,0,0,n));          break;      }    }  }  return 0;}