HDU 1394 Minimum Inversion Number（线段树+逆序数）

HDU 1394 Minimum Inversion Number（线段树+逆序数）

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

听过学长的讲课之后，还是不懂，后来发现自己把题意理解错了，orz。果然是被dp洗脑了，以为是求最大逆序数……研究了下别人的代码，发现其实很简单，和之前做的题差不多……代码如下~

#include<iostream>
#include<stdio.h>
#include<queue>
#include<stack>
#include<algorithm>
#include<string.h>
#include<string>
#include<math.h>
using namespace std;
struct node
{
int left,right,sum;
}tree[20000];
void build(int id,int l,int r)
{
tree[id].left=l;
tree[id].right=r;
tree[id].sum=0;
if(l==r)
return;
int mid=(tree[id].left+tree[id].right)/2;
build(2*id,l,mid);
build(2*id+1,mid+1,r);
}//建树过程不用多说了~
void update(int id,int val)
{
if(tree[id].left==val&&tree[id].right==val)
{
tree[id].sum=1;//当到达输入数字位置时sum变为1；
return;
}
int mid=(tree[id].left+tree[id].right)/2;
if(val<=mid)
update(id*2,val);//递归更新数据
else
update(id*2+1,val);
tree[id].sum=tree[id*2].sum+tree[id*2+1].sum;//父节点为左右儿子的和
}
int query(int id,int l,int r)
{
if(tree[id].left>=l&&tree[id].right<=r)
{
return tree[id].sum;
}
else
{
int mid=(tree[id].left+tree[id].right)/2;
int ans=0;
if(l<=mid) ans+=query(id*2,l,r);
if(mid<r)  ans+=query(id*2+1,l,r);
return ans;
}
}//参照了线段树区间和的模板
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
build(1,0,n-1);
int ans=0;
int a[5005];
for(int i=0;i<n;i++)
{
scanf("%d",&a[i]);
ans+=query(1,a[i]+1,n-1);//每次输入以后求一次该数的逆序数，最终和即为所有逆序数的和
update(1,a[i]);//更新树
}
int Min=ans;
for(int i=0;i<n;i++)
{
ans=ans-a[i]+(n-1-a[i]);//删除第i个数，会减少a[i]个逆序数，因为只有n-1个数，所以增加n-1-a[i]个逆序数；
Min=min(Min,ans);
}
printf("%d\n",Min);
}
return 0;
}