Minimum Inversion Number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13036 Accepted Submission(s): 7968
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
题目意思：
给一个n，然后一个长度为n的数组，每个元素小于n且不重复，每次操作把首位元素放到末尾，求n次操作中最小的逆序对数目。
思路：
先求一下原始数组的逆序对数目，每次把首位放末尾的时候，num（逆序对数目）加上大于a[i]的，然后减去小于a[i]的。
代码：
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <vector>
#include <queue>
#include <cmath>
#include <set>
using namespace std; #define N 5005
#define ll root<<1
#define rr root<<1|1
#define mid (a[root].l+a[root].r)/2 int max(int x,int y){return x>y?x:y;}
int min(int x,int y){return x<y?x:y;}
int abs(int x,int y){return x<?-x:x;} int n;
int b[N];
int a[N]; int lowbit(int x){
return x&(-x);
} void solve(int val,int x){
while(x<=n){
a[x]+=val;
x+=lowbit(x);
}
} int sum(int x){
int ans=;
while(x>){
ans+=a[x];
x-=lowbit(x);
}
return ans;
} main()
{
int t, i, j, k;
while(scanf("%d",&n)==){
memset(a,,sizeof(a));
int ans=;
for(i=;i<=n;i++) {
scanf("%d",&b[i]);
b[i]++;
ans+=sum(n)-sum(b[i]);
solve(,b[i]);
}
int num=ans;
for(i=;i<=n;i++){
num+=n-b[i]-b[i]+;//sum(n)-sum(b[i])-sum(b[i]-1);
ans=min(ans,num);
}
printf("%d\n",ans);
}
}