HDU 1394 calculates the number of reverse orders

First paste a TLE code...

# Include <iostream> # include <cstdio> # define MAXN 11111 using namespace std; int gt [MAXN <2], st [MAXN <2]; int date [MAXN]; void pushUpmax (int rt) {gt [rt] = max (gt [rt <1], gt [rt <1 | 1]);} void pushUpmin (int rt) {st [rt] = min (st [rt <1], st [rt <1 | 1]);} void build (int l, int r, int rt) {if (l = r) {gt [rt] = st [rt] = date [l]; return ;} int m = (l + r)/2; build (l, m, rt <1); build (m + 1, r, rt <1 | 1 ); pushUpmax (rt); pushUpmin (Rt);} int querygt (int v, int l, int r, int rt) // find the number of nodes smaller than v in the [l, r] range on the maximum tree. {If (l = r) return 0; if (gt [rt] <v) return r-l + 1; int m = (l + r)/2; int ret = 0; ret + = querygt (v, l, m, rt <1); ret + = querygt (v, m + 1, r, rt <1 | 1); return ret;} int queryst (int v, int l, int r, int rt) // search for [l, r] number of nodes greater than v in the interval {if (l = r) return 0; if (gt [rt]> v) return r-l + 1; int m = (l + r)/2; int ret = 0; ret + = queryst (v, l, m, rt <1); ret + = queryst (v, m + 1, r, rt <1 | 1); return ret;} int main () {int N; while (scanf ("% d", & N )! = EOF) {for (int I = 1; I <= N; I ++) {scanf ("% d", & date [I]); date [I + N] = date [I];} build (1, 2 * N, 1); int ans = 0; for (int I = 1; I <= N; I ++) for (int j = I + 1; j <= N; j ++) if (date [I]> date [j]) ans ++; for (int I = 1; I <= N; I ++) ans = min (ans, ans + queryst (date [I], I + 1, I + N-1, 1)-querygt (date [I], I + 1, I + N-1, 1); printf ("% d \ n", ans);} return 0 ;}

Although it was me, it was my attempt. Although TLE is... helpless ..

First, define the number of reverse orders:

The number of Ai> Aj in a string of numbers. Conversely, when I> j, the number of Ai <Aj.

Generally speaking, it is the number of numbers that are larger than Ai before.

Here is a brief description of my formula.

| ...... | AB | ...... |

Assuming that AB is the two adjacent numbers in the string, how does the reverse order change the positions of the two numbers?

| ...... | BA | ...... |

If (A> B) r --;

If (B> A) r ++;

Obviously, the transposition of AB is irrelevant to the following two things: | ...... |.

So what about A |... | transpose |... |?

Let's take a step-by-step look...

ABCDEFG

BACDEFG

BCADEFG

BCDAEFG

BCDEAFG

BCDEFAG

BCDEFGA

In this way, move A to the end.

How does the number of reverse Orders change in this process?

We can simply infer that the number of reverse orders added by A is greater than that of A in [B, G. The number of descending orders reduced by A is smaller than that of.

So... R + = [B, G] is greater than-[B, G] of A and smaller than.

R is the original reverse order number.

This is the origin of the above code...

Unfortunately, TLE...

The question has a special nature. Number has N for [0, N-1] each has.

So almost move Ai to the end, where there are (N-Ai-1) More than Ai, less than Ai there are Ai, so

R + = (N-Ai-1)-Ai;

This is the formula for a change.

So...

Let's do it with brute force...

#include<iostream>using namespace std;int main(){ int N; while( scanf( "%d",&N )!=EOF ) {    int date[5555],ans=0;    for( int i=0;i<N;i++ )    scanf( "%d",&date[i] );    for( int i=0;i<N;i++ )    for( int j=0;j<i;j++ )    if( date[j]>date[i] )    ans++;    int temp=ans;    for( int i=0;i<N;i++ )    {   temp=temp-date[i]+(N-date[i]-1);    ans=min( ans,temp );   }    printf("%d\n",ans );  }  return 0;}

Tree array:

#include<iostream>using namespace std;int tree[5555],N;void update( int pt,int v ){  while( pt<=N )  {  tree[pt]+=v;  pt+=pt&(-pt); }}int find( int pt ){ int ret=0; while( pt ) {     ret+=tree[pt];     pt-=pt&(-pt);    } return ret;}int main(){ int date[5555]; while( scanf("%d",&N)!=EOF ) {    memset( tree,0,sizeof(tree) );    for( int i=0;i<N;i++ )    {    scanf( "%d",&date[i] );    date[i]++;   }    int ans=0;   for( int i=N-1;i>=0;i-- )    {   ans+=find(date[i]);   update(date[i],1);   }   printf( "%d\n",ans );    int temp=ans;   for( int i=0;i<N;i++ )   {   date[i]--;   temp=temp-date[i]+(N-date[i]-1);   ans=min(temp,ans);      }      printf( "%d\n",ans );  }  return 0;}