hdu 1394 Minimum Inversion Number

[align=left]Problem Description[/align]
The inversion number of a given number sequence a1, a2, ..., an is the number of pairs (ai, aj) that satisfy i < j and ai > aj.

For a given sequence of numbers a1, a2, ..., an, if we move the first m >= 0 numbers to the end of the seqence, we will obtain another sequence. There are totally n such sequences as the following:

a1, a2, ..., an-1, an (where m = 0 - the initial seqence)

a2, a3, ..., an, a1 (where m = 1)

a3, a4, ..., an, a1, a2 (where m = 2)

...

an, a1, a2, ..., an-1 (where m = n-1)

You are asked to write a program to find the minimum inversion number out of the above sequences.

[align=left]Input[/align]
The input consists of a number of test cases. Each case consists of two lines: the first line contains a positive integer n (n <= 5000); the next line contains a permutation of the n integers from 0 to n-1.

[align=left]Output[/align]
For each case, output the minimum inversion number on a single line.

[align=left]Sample Input[/align]

10
1 3 6 9 0 8 5 7 4 2

[align=left]Sample Output[/align]

16

我采用线段树解决此问题；每插入一个结点，就检查之前是否有更大的数插入，以此统计逆序对；注意到数字是0到n的一个序列，所以每次将队头的数字移到队尾，逆序对的增减可以找出递推式；

[code]#include<iostream>
#include<cstdlib>
#include<string>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<stack>
#include<queue>
#include<iomanip>
#include<map>
#define pi 3.14159265358979323846
using namespace std;
int s[5001];
int tree[20001];
void build(int p,int l,int r)
{
if(l==r) {tree[p]=0;return;}
int mid=(l+r)/2;
build(p*2,l,mid);
build(p*2+1,mid+1,r);
tree[p]=0;
}
void change(int p,int l,int r,int x,int num)
{
if(l==r) {tree[p]+=num;return;}
int mid=(l+r)/2;
if(x<=mid) change(p*2,l,mid,x,num);
else change(p*2+1,mid+1,r,x,num);
tree[p]=tree[p*2]+tree[p*2+1];
}
int find(int p,int l,int r,int x,int y)
{
if(x<=l&&r<=y) return tree[p];
int mid=(l+r)/2;
if(y<=mid) return find(p*2,l,mid,x,y);
if(x>mid) return find(p*2+1,mid+1,r,x,y);
return (find(p*2,l,mid,x,mid)+find(p*2+1,mid+1,r,mid+1,y));
}
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
build(1,1,5000);
int ans=0;
for(int i=1;i<=n;++i)
{
scanf("%d",&s[i]);
ans+=find(1,1,5000,s[i]+1,5000);//统计逆序对
change(1,1,5000,s[i]+1,1);
}
int smin=ans;
for(int i=1;i<=n;++i)
{
ans=ans+n-2*s[i]-1;//递推式
if(ans<smin)
smin=ans;
}
printf("%d\n",smin);
}
return 0;
}

[/code]