Minimum Inversion Number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18375 Accepted Submission(s): 11156
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.
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
Sample Output
16
Author
CHEN, Gaoli
Source
ZOJ Monthly, January 2003
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题解：问你不断地将开头的数放到结尾。求其中最小的逆序数。
线段树求出最初的逆序数。复杂度O（nlogn），然后O(n)查询其他解。
AC代码：
#pragma comment(linker, "/STACK:102400000,102400000")
//#include<bits/stdc++.h>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <vector>
#include <map>
#include <cmath>
#include <queue>
#include <set>
#include <bitset>
#include <iomanip>
#include <list>
#include <stack>
#include <utility>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef vector<int> vi;
const double eps = 1e-8;
const int INF = 1e9+7;
const ll inf =(1LL<<62) ;
const int MOD = 1e9 + 7;
const ll mod = (1LL<<32);
const int N =1e6+6;
const int M=100010;
const int maxn=55555;
#define mst(a) memset(a, 0, sizeof(a))
#define M_P(x,y) make_pair(x,y)
#define in freopen("in.txt","r",stdin)
#define rep(i,j,k) for (int i = j; i <= k; i++)
#define per(i,j,k) for (int i = j; i >= k; i--)
#define lson l , mid , rt << 1
#define rson mid + 1 , r , rt << 1 | 1
const int lowbit(int x) { return x&-x; }
//const int lowbit(int x) { return ((x)&((x)^((x)-1))); }
int read(){ int v = 0, f = 1;char c =getchar();
while( c < 48 || 57 < c ){if(c=='-') f = -1;c = getchar();}
while(48 <= c && c <= 57) v = v*10+c-48, c = getchar();
return v*f;}
int sum[maxn<<2];
int x[maxn];
void pushup(int rt) //把当前结点的信息更新到父结点
{
//线段树是用数组来模拟树形结构
//对于每一个节点rt,左子节点为 2*rt (一般写作rt<<1)右子节点为 2*rt+1（一般写作rt<<1|1）
sum[rt] = sum[rt<<1] + sum[rt<<1|1];
}
void build(int l,int r,int rt)
{
sum[rt] = 0;
if(l==r) return ;
int mid=(l+r)>>1;
build(lson);//递归构造左子树
build(rson);//递归构造右子树
pushup(rt); //更新求和
}
void update(int p, int l, int r, int rt)//单点增
{
if(l==r){
sum[rt] ++; return ;
}
int mid=(l+r)>>1;
if(p<=mid) update(p,lson);
else update(p,rson);
pushup(rt);
}
int query(int L,int R,int l,int r,int rt) //区间求逆序数
{
if(L <= l && r <= R){
return sum[rt];
}
int mid = (l+r)>>1;
int ans = 0;
if(L <= mid) ans+=query(L,R,lson);
if(R > mid) ans+=query(L,R,rson);
return ans;
}
int main()
{
int n;
while(~scanf("%d",&n))
{
build(0,n-1,1);
int sum = 0;
for(int i=0; i<n; i++)
{
scanf("%d",&x[i]);
sum+=query(x[i],n-1,0,n-1,1);
update(x[i],0,n-1,1);
}
int ans = sum;
for(int i=0;i<n;i++)
{
sum=sum-x[i]; //减少了的逆序数
sum+=n-x[i]-1; //增加了的逆序数
ans=min(ans,sum);
}
cout<<ans<<endl;
}
return 0;
}
