hdu Minimum Inversion Number

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5210    Accepted Submission(s): 3190

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

ZOJ Monthly, January 2003

 

Recommend

Ignatius.L

 

分析：线段树：单点更新，成段求和。模拟次序。

#include<cstdio>
#include<cstring>
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
int sum[1 << 14];
int x[5010];

inline int max(int a, int b) {
    return a > b ? a : b;
}

void pushup(int rt) {
    sum[rt] = (sum[rt << 1] + sum[rt << 1 | 1]);
}

void build(int l, int r, int rt) {
    if (l == r) {
        return;
    }
    int m = (l + r) >> 1;
    build(lson);
    build(rson);
}

void update(int p, int add, int l, int r, int rt) {
    if (l == r) {
        sum[rt] += add;
        return;
    }
    int m = (l + r) >> 1;
    if (p <= m)
        update(p, add, lson);
    else
        update(p, add, 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 ans = 0;
    if (L <= m)
        ans = (ans + query(L, R, lson));
    if (R > m)
        ans = (ans + query(L, R, rson));
    return ans;
}

int main() {
    int n, i, j, ans, min;
    while (scanf("%d", &n) != EOF) {
        memset(sum, 0, sizeof (sum));
        ans = 0;
        build(0, n - 1, 1);
        for (i = 0; i < n; ++i) {
            scanf("%d", &x[i]);
            ans += query(x[i]+1, n - 1, 0, n - 1, 1);
            update(x[i], 1, 0, n - 1, 1);
        }
        min = ans;
        for (i = 0; i < n - 1; ++i) {
            ans = ans - x[i] - x[i] + n - 1;
            if (ans < min)
                min = ans;
        }
        printf("%d\n", min);
    }
    return 0;
}