HDU 1394 Minimum Inversion Number（归并排序 线段树 逆序数）
http://acm.hdu.edu.cn/showproblem.php?pid=1394
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
题意：
有n个从0到n 的数 按照给定次序排列 每次只能将最前的一个数移到末尾 算出每种情况的逆序数 并求出逆序数最小值
思路：
归并排序或线段树
首先 我们把当前情况下的逆序数求出
然后循环n次 每次最前一个往最后移 则 ans=ans+ (n-1+a[i])-a[i]
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
#include<algorithm>
using namespace std;
int num[5000+10],a[5000+10],t[5000+10];
//int vis[5000+10];
int ans;
void Merge(int x,int m,int y)
{
int p=x,q=m+1,i=x;
while(p<=m||q<=y)
{
if(q>y||(p<=m&&a[p]<=a[q])) t[i++]=a[p++];
else
{
t[i++]=a[q++];
ans+=m-p+1;
}
}
for(i=x;i<=y;i++) a[i]=t[i];
}
void mergesort(int x,int y)
{
if(x!=y)
{
int mid=(x+y)/2;
mergesort(x,mid);
mergesort(mid+1,y);
Merge(x,mid,y);
}
}
int main()
{
int n,i,j,minn;
while(scanf("%d",&n)!=EOF)
{
ans=0;
for(i=1;i<=n;i++)
{scanf("%d",&a[i]); num[i]=a[i];}
mergesort(1,n);
minn=ans;
for(i=1;i<n;i++)
{
ans=ans+n-1-2*num[i];
if(ans<minn) minn=ans;
}
cout<<minn<<endl;
}
return 0;
}
线段树的版本
对于初始序列
每次加入一个数就对其标记 并且查询是否有比他大的数已经被加入 从而求出逆序数
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#define mem(a,b) memset(a,b,sizeof(a))
#define ll __int64
#define MAXN 1000
#define INF 0x7ffffff
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
int sum[50000];
int a[5000+10];
void pushup(int rt)
{
sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}
void build(int l,int r,int rt)
{
if(l==r)
{
sum[rt]=0; return ;
}
int m=(l+r)>>1;
build(lson);
build(rson);
pushup(rt);
}
int getnum(int L,int R,int l,int r,int rt)
{
if(L<=l&&r<=R)
{
return sum[rt];
}
int ret=0;
int m=(l+r)>>1;
if(L<=m) ret+=getnum(L,R,lson);
if(R>m) ret+=getnum(L,R,rson);
return ret;
}
void update(int num,int l,int r,int rt)
{
if(l==r&&l==num)
{
sum[rt]=1;
return ;
}
int m=(l+r)>>1;
if(num<=m) update(num,lson);
else update(num,rson);
pushup(rt);
}
int main()
{
int n;
int i,j;
int ans,minn;
while(scanf("%d",&n)!=EOF)
{
ans=0;
build(0,n-1,1);
for(i=0;i<n;i++)
{
scanf("%d",&a[i]);
ans+=getnum(a[i],n-1,0,n-1,1);
update(a[i],0,n-1,1);
}
minn=ans;
for(i=0;i<n-1;i++)
{
ans=ans-a[i]*2+n-1;
if(minn>ans) minn=ans;
}
cout<<minn<<endl;
}
return 0;
}