HDU 1394 Minimum Inversion Number （线段数）

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13951    Accepted Submission(s): 8519

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

 

题意:求Inversion后的最小逆序数
思路:用O(nlogn)复杂度求出最初逆序数后,就可以用O(1)的复杂度分别递推出其他解
线段树功能:update:单点增减 query:区间求和

（初学线段树。。借某位不知名大神的模板）

#include <stdio.h>
#include <string.h>
#include <algorithm>
#define INF 0x3f3f3f3f
#define lson l, m, rt << 1
#define rson m+1, r, rt << 1 | 1
using namespace std;

int sum[5010 << 2];
int x[5010];
void build(int l, int r, int rt)
{
    memset(sum, 0, sizeof(sum));
}
int query(int L, int R, int l, int r, int rt)
{
    if(L <= l && r <= R)
        return sum[rt];
    int m = (r + l) >> 1;
    int res = 0;
    if(L <= m) res += query(L, R, lson);
    if(R > m) res += query(L, R, rson);
    return res;
}
void pushup(int rt)
{
    sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
}
void update(int p, int add, int l, int r, int rt)
{
    if(l == r)
    {
        sum[rt] += add;
        return;
    }
    int m = (r + l) >> 1;
    if(p <= m) update(p, add, lson);
    else update(p, add, rson);
    pushup(rt);
}
int main()
{
    int n;
    while(~scanf("%d", &n))
    {
        build(0, n-1, 1);
        int sum = 0;
        for(int i=0; i<n; i++)
        {
            scanf("%d", &x[i]);
            sum += query(x[i], n-1, 0, n-1, 1);
            update(x[i], 1, 0, n-1, 1);
        }
        int res = sum;
        for(int i=0; i<n; i++)
        {
            sum = sum + (n - x[i] - 1) - x[i];
            res = min(res, sum);
        }
        printf("%d\n", res);
    }
    return 0;
}