HDU 1394 Minimum Inversion Number

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 21957    Accepted Submission(s): 13098

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

 

思路：线段树区间最值，单点增减，求最小逆序数，先求出一个，然后可以发现公式，进而推出其他的解

 

实现代码：

#include<bits/stdc++.h>
#include<cstdio>
#include<algorithm>
using namespace std;

#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
const int M = 5555;
int sum[M<<2];
int n;
void pushup(int rt){
    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 m = (l+r) >> 1;
    build(lson);
    build(rson);
}

void update(int p,int l,int r,int rt){
    if(l == r){
        sum[rt] ++;
        return ;
    }
    int m = (l + r) >> 1;
    if(p <= m) update(p,lson);
    else update(p,rson);
    pushup(rt);
}

int query(int L,int R,int l,int r,int rt){
    if(L <= l && r <= R){
        return sum[rt];
    }
    int m = (l + r) >> 1;
    int ret = 0;
    if(L <= m) ret += query(L,R,lson);
    if(R>m) ret += query(L,R,rson);
    return ret;
}
int a[M];
int main(){
    int sum1;
    while(scanf("%d",&n)!=EOF){
         build(0,n-1,1);
         int sum1 = 0;
         for(int i = 0;i < n;i++){
            scanf("%d",&a[i]);
            sum1 += query(a[i]+1,n-1,0,n-1,1);
            update(a[i]+1,0,n-1,1);
         }
         int ans = sum1;
         for(int i = 0;i < n; i++){
            sum1 = sum1 - a[i] + n - a[i] - 1;
            ans = min(ans,sum1);
         }
         printf("%d
",ans);
    }
    return 0;
}