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
本题就是求在a1,a2,…an的不断交换排列中，最小的逆序数之和，正解为按index建树，统计每个排列的逆序数之和。
///@zhangxiaoyu
///2015/8/1
///此题正解为线段树，然而建树失败
///由于数据小，暴力过了此题，暴力代码如下，以后把线段树的代码贴上
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<vector>
#include<stack>
#include<map>
#include<queue>
#include<iostream>
using namespace std;
int a[5010];
int n;
int solve1(int num)///调整一个排列后，比交换的数小的
{
return num;
}
int solve2(int num)///调整后，num之前的逆序数的变化
{
return n-num-1;
}
int main()
{
while(~scanf("%d",&n))
{
int cnt=0;
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
for(int j=1;j<i;j++)
{
if(a[j]>a[i])
cnt++;
}
}
int ans=cnt;
for(int i=1;i<=n;i++)
{
cnt=cnt-solve1(a[i])+solve2(a[i]);
if(cnt<ans)
ans=cnt;
}
printf("%d\n",ans);
}
return 0;
}
///线段树版
#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;
}
