Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4904    Accepted Submission(s): 2991


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
Recommend Ignatius.L 首先，求逆序數對的思路： 1.得到整個數列後，從前往後掃，統計比a[i]小的，在a[i]後面的有多少個 這樣做的話，應該是隻有n2的暴力作法，沒想到更好的方法 2.統計a[i]前面的，且比它大的數 這樣做的話，就可以利用輸入的時效性，每輸入一個數，就把這個數的num[i]值加1， 然後統計比這個數大的數的num和, 因為這裡的和一定是在這個數列中比a[i]大，且在它前面出現的數之和， 然後把把這個和加到總逆序數sum裡。 這樣做的話直接的暴力作法依然是n2，但是， 我們可以在，統計比這個數大的數的num和這一步進行優化，利用線段樹求區間域值的複雜度是logn， 所以總體複雜度就降到了nlogn。 再來看這道題，求得初始數列的逆序數後，再求其他排列的逆序數有一個規律，就是 sum = sum + (n - 1 - a[i]) - a[i]; 這個自行驗證吧，相信很容易得出 最後，拓展一下，如果要求正序數怎麼辦？很簡單，無非是大小調一下 再問，如果要求滿足i<j<k,且a[i]>a[j]>a[k]的數對總數怎麼辦？ 可以從中間的這個數入手，統計a[i]>a[j]的對數m，以及a[j]>a[k]的對數n，m*n就是。。。 要求a[i]>a[j]的個數還是一樣的，那麼a[j]>a[k]的個數呢？ 兩種思路： 1.得到a[i]>a[j]的對數後，將數列倒過來後再求a[j]<a[k]的對數 2.更簡單的做法是，找到規律發現，n = 整個數列中比a[j]小的數 — 在a[j]前面已經出現的比a[j]小的數的個數 即（假設數列是從1開始的） n = (a[j] -1) - (j - 1 - m ) 這道題程式碼

#include <stdio.h>
#include <algorithm>

using namespace std;

int a[5005];
struct Node{
    int l,r,num;
}tree[50000];

void Build(int n,int x,int y){
    tree[n].l = x;
    tree[n].r = y;
    tree[n].num = 0;
    if(x == y){
        return;
    }
    int mid = (x + y) / 2;
    Build(2*n,x,mid);
    Build(2*n+1,mid+1,y);
}

void Modify(int n,int x){
    int l = tree[n].l;
    int r = tree[n].r;
    int mid = (l + r) / 2;
    if(x == l && x == r){
        tree[n].num = 1;
        return;
    }
    if(x <= mid)    Modify(2*n,x);
    else            Modify(2*n+1,x);
    tree[n].num = tree[2*n].num + tree[2*n+1].num;
}

int Query(int n,int x,int y){
    int l = tree[n].l;
    int r = tree[n].r;
    int mid = (l + r) / 2;
    int ans = 0;;
    if(x == l && y == r)
        return tree[n].num;
    if(x <= mid)   ans += Query(2*n,x,min(mid,y));
    if(y > mid)    ans += Query(2*n+1,max(mid+1,x),y);
    return ans;
}
int main(){
    int n,sum,ans;
    int i,j;

    while(scanf("%d",&n) != EOF){
        sum = 0;
        Build(1,0,n);
        for(i = 1;i <= n;i++){
            scanf("%d",&a[i]);
            Modify(1,a[i]);
            sum += Query(1,a[i]+1,n);
        }
        ans = sum;
        for(i = 1;i < n;i++){
            sum = sum + (n - 1 - a[i]) - a[i];
            if(sum < ans)
                ans = sum;
        }
        printf("%d\n",ans);
    }
}