http://acm.hdu.edu.cn/showproblem.php?pid=1394
Minimum Inversion Number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 8205 Accepted Submission(s): 5041
Problem Description
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.
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.
Input
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.
Output
For each case, output the minimum inversion number on a single line.
Sample Input
10
1 3 6 9 0 8 5 7 4 2
Sample Output
16
#include<stdio.h>
#define maxn 500004
struct node
{
int left,right;
int sum;
};
int val[maxn];
int n;
node tree[*maxn];
void build(int left,int right,int i)
{
tree[i].left =left;
tree[i].right =right;
tree[i].sum =;
if(tree[i].left ==tree[i].right )
return ;
build(left,(left+right)/,*i);
build((left+right)/+,right,*i+);
}
void update(int r,int j)
{
tree[r].sum++;
if(tree[r].left==tree[r].right)
return ;
int mid=(tree[r].left +tree[r].right )/;
if(j<=mid)
update(r*,j);
if(j>mid)
update(r*+,j);
}
int getsum(int p, int left, int right)
{
if (left > right)
return ;
if (tree[p].left == left && tree[p].right == right)
return tree[p].sum;
int m = (tree[p].left + tree[p].right) / ;
if (right <= m)
return getsum(p*, left, right);
else if (left > m)
return getsum(p*+, left, right);
else return getsum(p*, left, m) + getsum(p*+, m + , right);
}
int main()
{
while (~scanf("%d",&n))
{
build(,n - ,);
int i,sum =,ans;
for (i = ;i<= n; i++)
{
scanf("%d", &val[i]);
sum += getsum(, val[i], n - );
update(, val[i]);
}
ans = sum;
for (i = ; i <= n; i++)
{
sum = sum + (n - val[i] - ) - val[i];
ans = sum < ans ? sum : ans;
}
printf("%d\n", ans);
}
return ;
}