Luo gu 2657 bow a clan

Luo gu 2657 bow a clan

Address: http://www.luogu.org/problem/show?pid=2657

Title Description

A group of young people lined up to chat with each other on their mobile phones.
Each person's cell phone has a signal receiving indicator, and the first person's receive indicator is set to V[i].
If the person in X[i] is to chat with the person who is in XJ, then the two-member signal emission intensity is abs (X[I]-X[J]) *max (V[i],v[j]).
Now we want to know the sum of the intensity of the signals emitted by these people in all pairs.

Input/output format

Input format:

The first line is an integer n, followed by n lines, two integers per line v[i] and X[i].

Output format:

The total signal emission intensity of all pairs.

Input/Output sample

Input Sample # #:

4

3 1

2 5

2 6

4 3

Sample # # of output:

57

Description

For 40% of data, n<=5,000
For 100% of data, n<=100,0001≤x[i]≤20,000
Note: There may be two people in the same position
The answer is within the range of int64 or long long

Ideas

A tree-like array.

First sort the people in order of V from large to small, which is equivalent to eliminating the impact of V, the person previously considered v must be less than or equal to the current v.

Second consider the ABS (X[i]-x[j]) part, just beginning to think is the prefix and, but note that

ABS (X-X1) +abs (x-x2)!=abs (2*x-(X1+X2))

Here, NUML represents the number of previous x smaller than the current X, SUML represents the sum of x, a similar definition SUMR,NUMR the previous x is smaller than the current x. The current

ans=v* (x*numl-suml+sumr-numr*x)

The numl suml here can be obtained by bit in the time of O (Logn). The total time complexity is O (NLOGN).

It is important to note that starting with X or starting a query from X-1 is no different from the subject.

Code

1#include <cstdio>2#include <iostream>3#include <cstring>4#include <algorithm>5 #definefor (A,B,C) for (int a= (b); a<= (c); a++)6 using namespacestd;7 8typedefLong LongLL;9 Const intMAXN =100000+Ten, maxm=20000+Ten;Ten structnode{ One     intx,v; A     BOOL operator< (ConstNODE&AMP;RHS)Const{ -         returnv<rhs.v; -     } the }NODES[MAXN]; - intN,max; - ////bit - intLowbit (intx) {returnx& (-x); } + voidADD (LL *c,intXintv) { -      while(x<=Max) { +C[x]+=v; x+=lowbit (x); A     } at } -ll query (ll *c,intx) { -LL res=0; -      while(x>0) { -RES+=C[X]; x-=lowbit (x); -     } in     returnRes; - } to LL NUM[MAXM],SUM[MAXM]; +  - ////read the intRead_int () { *     CharC=GetChar (); $      while(!isdigit (c)) c=GetChar ();Panax Notoginseng     intx=0; -      while(IsDigit (c)) { thex=x*Ten+c-'0'; +C=GetChar (); A     } the     returnx; + }  - intMain () $ { $n=read_int (); -for (I,1, N) { -nodes[i].v=read_int (); thenodes[i].x=read_int (); -max=Max (max,nodes[i].x);Wuyi     } theSort (nodes+1, nodes+n+1); -LL ans=0; Wufor (I,1, N) { -         intx=nodes[i].x,v=nodes[i].v; AboutLL Numl=query (num,x-1); $LL Suml=query (sum,x-1); -LL Numr=query (Num,max)-query (num,x); -LL Sumr=query (Sum,max)-query (sum,x); -Ans + = v* (sumr-x*numr+x*numl-suml); AADD (Num,x,1); + Add (sum,x,x); the     } -cout<<ans<<"\ n"; $     return 0; the}

Luo gu 2657 bow a clan