POJ 1990 Moofest Tree-like array

The problem is that I saw the big Baishushu-shaped array after the first question of the brush, it is really difficult, so have to find the solution on the Internet, the practice of online is really wonderful. The main test instructions of this problem is n cattle, the cost of exchange between two cows is the maximum of their distance by the volume of both (that is, Max (V (i), V (j))), and then the total cost of all cattle 22 exchange. The first thing to think of is O (N2) of the violence enumerated, then I can only think of such a complexity, it is very curious how to catch up with the tree array edge? Then after looking at the other people's solution to marvel at its ideas.

In the blog http://www.cnblogs.com/Fatedayt/archive/2011/10/08/2202439.html very detailed, attached its main ideas:

has been described in detail, that is, the statistics of a cow I, by sequencing and preprocessing can be the original O (n) query down to O (Logn), is really very powerful, it is difficult to think. I combined with the other people on the Internet half copy half of the following code rewrite:

1#include <cstdio>2#include <cstring>3#include <algorithm>4 using namespacestd;5typedefLong LongLL;6 Const intmaxn=20020;7 8inline ll Lowbit (ll x) {returnx& (-x); }9 Ten structtreearray{ One LL C[MAXN], N; ATreearray (LL n=0): N (n) {memset (c,0,sizeof(c)); } - ll sum (ll x) { -LL ans=0; the          while(x) { -ans+=C[x]; -x-=lowbit (x); -         } +         returnans; -     } +     voidAdd (ll x, ll D) { A          while(x<=N) { atc[x]+=D; -x+=lowbit (x); -         } -     } -} Count (20003), Dist (20003); -  in structcow{ - LL v,x; to     BOOL operator< (ConstCow C2)Const { +         returnv<c2.v; -     } the } COW[MAXN]; *  $ intMain () {Panax Notoginseng     intn,i; -scanf"%d",&n); the      for(i=1; i<=n; ++i) +scanf"%lld%lld",&cow[i].v,&cow[i].x); ASort (cow+1, cow+n+1); theLL ans=0, alldist=0; +      for(i=1; i<=n; ++i) { -LL x =cow[i].x; $LL num =count.sum (x); $LL lessdist =dist.sum (x); -Ans + = cow[i].v* (num*x-lessdist+alldist-lessdist-(i-1-num) *x); -Count.add (x,1); the Dist.add (x,x); -Alldist + =x;Wuyi     } theprintf"%lld\n", ans); -     return 0; Wu}

Among them, Count.sum (x) indicates the number of cattle with a small amount of x (in the middle is a), Dist.sum (x) indicates the sum of the positions of the cattle (in B), which is smaller than x, and alldist is the total distance of the former I cattle (which can be considered as the standard for the origin), and the remaining variable names are After the code to feel the tree array is really too powerful, have to praise!

POJ 1990 Moofest Tree-like array