Minimum Inversion Number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 23159 Accepted Submission(s): 13765
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
10 1 3 6 9 0 8 5 7 4 2
Sample Output
16
Author
CHEN, Gaoli
Source
题意：求数组中一共有多少组逆序数对，数组可以变化，每次把第一个数放在数组最后面，求其中出现的最少的逆序数对。
思路：先考虑原来的数组，例如案例一。数组为：1 3 6 9 0 8 5 7 4 2
当数组中元素只有1时，逆序对为0
当数组中元素只有1和3时，3前面的数小于3，逆序对为0
当数组中元素只有1、3、6时，6前面的数小于6，逆序对为0
当数组中元素只有1、3、6、9时，9前面的数小于9，逆序对为0
当数组中元素只有1、3、6、9、0时，0前面存在1，3，6，9，逆序对为4
当数组中元素只有1、3、6、9、0、8时，8前面存在9，逆序对为4+1
.。。。
因此我们可以每次插入一个数a[i]、然后统计之前插入且比a[i]大的数有多少个。
使用线段树可以实现区间的查询操作。
建好树后，每次逆序对的值加上树[a[i]+1,n-1]的值，然后将当前的a[i]插入到树中。
对于数组的每次把第一个数放到最后的操作：
比a[i]小的数有a[i]个，比a[i]大的数有n-a[i]-1个，因此
逆序对的值+=(n-a[i]-1)-a[i];
AC代码：
#include <stdio.h>
#include <algorithm>
typedef long long ll;
const int MAX_N = 1e6 + 100;
using namespace std;
int n,m;
int a[MAX_N];
struct node{
int l,r,lazy;
ll sum;
void update(ll x){
sum+=1ll*(r-l+1)*x;
lazy+=x;
}
}tree[MAX_N>>2];
void push_up(int x){
tree[x].sum=tree[x<<1].sum+tree[x<<1|1].sum;
}
void push_down(int x){
int lazyvalue=tree[x].lazy;
if(lazyvalue>0){
tree[x<<1].update(lazyvalue);
tree[x<<1|1].update(lazyvalue);
tree[x].lazy=0;
}
}
void build(int x,int l, int r){
tree[x].l=l;tree[x].r=r;
tree[x].sum=0;tree[x].lazy=0;
if(l==r){
tree[x].sum=0;
return ;
}
int mid=(l+r)>>1;
build(x<<1,l,mid);
build(x<<1|1,mid+1,r);
push_up(x);
}
void update(int x,int l,int r,ll val){
int L=tree[x].l;int R=tree[x].r;
int mid=(L+R)>>1;
if(l<=L && R<=r){
tree[x].update(val);
return ;
}
push_down(x);
if(l<=mid){
update(x<<1,l,r,val);
}
if(mid<r){
update(x<<1|1,l,r,val);
}
push_up(x);
}
ll query(int x,int l,int r){
int L=tree[x].l;int R=tree[x].r;
int mid=(L+R)>>1;
if(l<=L && R <=r){
return tree[x].sum;
}
push_down(x);
ll ans=0;
if(l<=mid){
ans+=query(x<<1,l,r);
}
if(mid<r){
ans+=query(x<<1|1,l,r);
}
return ans;
}
int main(){
while(~scanf("%d",&n)){
for(int i=0;i<n;i++){
scanf("%d",&a[i]);
}
build(1,0,n-1);
ll ans=0;
for(int i=0;i<n;i++){
ans+=query(1,a[i]+1,n-1);
update(1,a[i],a[i],1);
}
ll tmp=ans;
// printf("ans=%lld\n",tmp);
for(int i=0;i<n;i++){
tmp=tmp+n-a[i]-1-a[i];
// printf("%lld\n",tmp);
ans=min(tmp,ans);
}
printf("%lld\n",ans);
}
return 0;
}