hdu1394(Minimum Inversion Number)线段树

明知道是线段树，却写不出来，搞了半天，戳，没办法，最后还是得去看题解（有待于提高啊啊），想做道题还是难啊。

还是先贴题吧

HDU-1394 Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8324    Accepted Submission(s): 5115

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

思路：依次统计a[i]前面有多少个比它大，求和就是初始化序列的逆序数个数；所有的数都在第一个数之后，那么比第一个数大的个数为n-a[1] - 1,比a[1]小的数就是a[1](关键),每次把第一个数放到最后一个位置时，逆序数增加了n-a[1]-1-a[1]了，明白了这个就好做了。

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iomanip>
#include <set>
#include <map>
#include <vector>
#include <queue>
#define N 5005
using namespace std;
int a[N];
int segtree[N << 2];
void update(int ll, int rr, int l, int r, int p)
{
    if (ll <= l && rr >= r)
    {
        segtree[p]++;
        return;
    }
    int mid = (l + r) >> 1, pp = p << 1;
    if (mid >= rr)
        update(ll, rr, l, mid, pp);
    else
        if (mid < ll)
            update(ll, rr, mid + 1, r, pp + 1);
        else
        {
            update(ll, mid, l, mid, pp);
            update(mid + 1, rr, mid + 1, r, pp + 1);
        }    
    segtree[p] = segtree[pp] + segtree[pp + 1];    
} 
int query(int ll, int rr, int l, int r, int p)
{
    if (ll <= l && rr >= r)
        return segtree[p]++;
    int mid = (l + r) >> 1, pp = p << 1;
    if (mid >= rr)
        query(ll, rr, l, mid, pp);
    else
        if (mid < ll)
            query(ll, rr, mid + 1, r, pp + 1);
        else
            return query(ll, mid, l, mid, pp) + 
            query(mid + 1, rr, mid + 1, r, pp + 1);
}
int main()
{
    int n, i, ans, minans;
    while (~scanf("%d", &n))
    {
        memset(segtree, 0, sizeof(segtree));
        ans = 0;
        for (i = 1; i <= n; i++)
        {
            scanf("%d", &a[i]);
            update(a[i] + 1, a[i] + 1, 1, n, 1);//将所有数加1，除去0，线段树下标从1开始
            if (a[i] + 1 < n)//当a[i] = n，不操作 
                ans += query(a[i] + 2, n, 1, n, 1); 
        }
        minans = ans;
        for (i = 1; i < n; i++)
        {
            ans += n - a[i] * 2 -1;
            if (minans > ans)
                minans = ans;
        }
        printf("%d
", minans);
    }
    return 0;
}