Hdu 1394 Minimum Inversion Number

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 5180    Accepted Submission(s): 3173


[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

 

[align=left]Author[/align]
CHEN, Gaoli
 

[align=left]Source[/align]
ZOJ Monthly, January 2003

 

[align=left]Recommend[/align]
Ignatius.L
 

树状数组

#include<stdio.h>
#include<string.h>

#define M 5001

int c[M];
int num[M];

int lowbit(int x)
{
return x&(-x);
}

int sum(int pos)
{
int s=0;
while(pos>0)
{
s+=c[pos];
pos-=lowbit(pos);
}
return s;
}

void update(int pos,int n)
{
while(pos<=n)
{
c[pos]+=1;
pos+=lowbit(pos);
}
}

int main(void)
{
int n;
//    freopen("d:\\in.txt","r",stdin);
while(scanf("%d",&n)==1)
{
int i,s=0,tmp;
memset(c,0,sizeof(c));
for(i=1;i<=n;i++)
{
scanf("%d",&num[i]);
}
for(i=n;i>=1;i--)              //逆序处理！！！！
{
s+=sum(num[i]);
//            printf("%d\n",s);
update(num[i]+1,n);
}
tmp=s;
for(i=1;i<=n;i++)
{
s+=n-num[i]-num[i]-1;
if(s<tmp)
tmp=s;
}
printf("%d\n",tmp);
}
return 0;
}