hdu 1394 Minimum Inversion Number

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

---

　　题目大意 求循环排列最小逆序对数。

　　老是刷cf，不做暑假作业估计会被教练鄙视，所以还是做做暑假作业。

　　先用各种方法算出原始序列的逆序对数。

　　显然你可直接算出把开头的一个数挪到序列后面增加的逆序对数。(后面有多少个比它大减去有多少个比它小)

　　于是这道题就水完了。

Code

/**
 * hdu
 * Problem#1394
 * Accepted
 * Time: 62ms
 * Memory: 1680k
 */
#include <iostream>
#include <fstream>
#include <sstream>
#include <cstdio>
#include <ctime>
#include <cmath>
#include <cctype>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <map>
#include <set>
#include <list>
#include <stack>
#include <queue>
#include <vector>
#ifndef WIN32
#define Auto "%lld"
#else
#define Auto "%I64d"
#endif
using namespace std;
typedef bool boolean;
const signed int inf = (signed)((1u << 31) - 1);
const signed long long llf = (signed long long)((1ull << 61) - 1);
const double eps = 1e-9;
const int binary_limit = 256;
#define smin(a, b) a = min(a, b)
#define smax(a, b) a = max(a, b)
#define max3(a, b, c) max(a, max(b, c))
#define min3(a, b, c) min(a, min(b, c))
template<typename T>
inline boolean readInteger(T& u){
    char x;
    int aFlag = 1;
    while(!isdigit((x = getchar())) && x != '-' && x != -1);
    if(x == -1) {
        ungetc(x, stdin);    
        return false;
    }
    if(x == '-'){
        x = getchar();
        aFlag = -1;
    }
    for(u = x - '0'; isdigit((x = getchar())); u = (u << 1) + (u << 3) + x - '0');
    ungetc(x, stdin);
    u *= aFlag;
    return true;
}

#define lowbit(x) ((x) & (-x))

typedef class IndexedTree {
    public:
        int* lis;
        int s;
        IndexedTree() {        }
        IndexedTree(int s):s(s) {
            lis    = new int[(s + 1)];
            memset(lis, 0, sizeof(int) * (s + 1));
        }
        
        inline void add(int idx, int val) {
            for(; idx <= s; idx += lowbit(idx))
                lis[idx] += val;
        }
        
        inline int getSum(int idx) {
            int ret = 0;
            for(; idx; idx -= lowbit(idx))
                ret += lis[idx];
            return ret;
        }
}IndexedTree;

int n;
int* arr;
inline boolean init() {
    if(!readInteger(n))    return false;
    arr = new int[(n + 1)];
    for(int i = 1; i <= n; i++)
        readInteger(arr[i]), arr[i] += 1;
    return true;
}

IndexedTree it;
int ans, cmp;
inline void solve() {
    ans = 10000000, cmp = 0;
    it = IndexedTree(n);
    for(int i = 1; i <= n; i++) {
        it.add(arr[i], 1);
//        cout << 1;
        cmp += i - it.getSum(arr[i]);
    }
    ans = cmp;
    for(int i = 1; i < n; i++) {
        cmp -= arr[i] - 1, cmp += n - arr[i];
        smin(ans, cmp);
    }
    printf("%d
", ans);
}

inline void clear() {
    delete[] arr;
    delete[] it.lis;
}

int main() {
    while(init()) {
        solve();
        clear();
    }
    return 0;
}