HDU 1394 Minimum Inversion Number

http://acm.hdu.edu.cn/showproblem.php?pid=1394

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

代码：

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

const int maxn = 5010;
int N;
int c[maxn], a[maxn];

int lowbit(int x) {
    return x & (-x);
}

void add(int x, int val) {
    while(x <= N) {
        c[x] += val;
        x += lowbit(x);
    }
}

int getsum(int x) {
    int sum = 0;
    while(x > 0) {
        sum += c[x];
        x -= lowbit(x);
    }
    return sum;
}

int main() {
    while(~scanf("%d", &N)) {
        if(!N) break;
        memset(c, 0, sizeof(c));
        int ans = 0;
        for(int i = 1; i <= N; i ++) {
            scanf("%d", &a[i]);
            a[i] += 1;
            ans += getsum(N) - getsum(a[i]);
            add(a[i], 1);
        }


        int minn = ans;
        for(int i = 1; i <= N; i ++) {
            ans += N - a[i] + 1 - a[i];
            minn = min(minn, ans);
        }
        printf("%d
", minn);
    }
    return 0;
}

　　树状数组 先把 a 数组输进来的时候加一 （因为平时写的逆序数的模板都是 1 开始的 呜呜呜） 本来写了一波暴力 稳稳的 TLE 但是每次拿一个到最后然后得到的逆序数数量是有规律的

15h