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.
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
这题就是求逆序数题，给你一个数列，找出它的逆序数，然后通过以上变换分别求出变换后的逆序数，输出最小值就是答案了。
假设ans是初始逆序数，那么他的下一个数列的逆序数ans1=ans+n-2*a[0]-1;因为把a[0]移到后面逆序数就要减少a[0],也要加上n-a[0]-1;如果这看懂了，这题就不难了，数据不大，求逆序数用线段树，暴力什么都能解决。
暴力：
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int main()
{
int n,a[100000],i,j,ans1,ans2;
while (~scanf("%d",&n))
{
ans1=0;
for (i=0;i<n;i++) scanf("%d",&a[i]);
for (i=1;i<n;i++)
for (j=0;j<i;j++)
if (a[i]<a[j]) ans1++;
ans2=ans1;
for (i=0;i<n;i++)
{
ans1=ans1+n-2*a[i]-1;
ans2=min(ans1,ans2);
}
printf("%d\n",ans2);
}
return 0;
}
线段树
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int s[50000];
struct p
{
int x,y,su;
};p tree[1000000];
void build(int l,int r,int p)
{
tree[p].x=l;
tree[p].y=r;
tree[p].su=0;
if (l==r) return ;
int m=(l+r)/2;
build(l,m,2*p);
build(m+1,r,2*p+1);
return ;
}
int find(int l,int r,int p)
{
if (tree[p].x==l&&tree[p].y==r) return tree[p].su;
int m=(tree[p].x+tree[p].y)/2;
if (l>m) return find(l,r,2*p+1);
if (m>=r) return find(l,r,2*p);
return find(l,m,2*p)+find(m+1,r,2*p+1);
}
void un(int pos,int i,int p)
{
tree[p].su+=i;
if (tree[p].x==tree[p].y) return ;
int m=(tree[p].x+tree[p].y)/2;
if (pos<=m) un(pos,i,2*p);
else un(pos,i,2*p+1);
}
int main()
{
int n,i,ans,mn;
while (~scanf("%d",&n))
{
build(0,n-1,1);
ans=0;
for (i=0;i<n;i++) scanf("%d",&s[i]);
for (i=0;i<n;i++)
{
ans+=find(s[i],n-1,1);
un(s[i],1,1);
}
mn=ans;
for (i=0;i<n;i++)
{
ans=ans+n-2*s[i]-1;
mn=min(mn,ans);
}
printf("%d\n",mn);
}
return 0;
}