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 << ];
 void update(int ll, int rr, int l, int r, int p)
 {
     if (ll <= l && rr >= r)
     {
         segtree[p]++;
         return;
     }
     int mid = (l + r) >> , pp = p << ;
     if (mid >= rr)
         update(ll, rr, l, mid, pp);
     else
         if (mid < ll)
             update(ll, rr, mid + , r, pp + );
         else
         {
             update(ll, mid, l, mid, pp);
             update(mid + , rr, mid + , r, pp + );
         }
     segtree[p] = segtree[pp] + segtree[pp + ];
 }
 int query(int ll, int rr, int l, int r, int p)
 {
     if (ll <= l && rr >= r)
         return segtree[p]++;
     int mid = (l + r) >> , pp = p << ;
     if (mid >= rr)
         query(ll, rr, l, mid, pp);
     else
         if (mid < ll)
             query(ll, rr, mid + , r, pp + );
         else
             return query(ll, mid, l, mid, pp) +
             query(mid + , rr, mid + , r, pp + );
 }
 int main()
 {
     int n, i, ans, minans;
     while (~scanf("%d", &n))
     {
         memset(segtree, , sizeof(segtree));
         ans = ;
         for (i = ; i <= n; i++)
         {
             scanf("%d", &a[i]);
             update(a[i] + , a[i] + , , n, );//将所有数加1，除去0，线段树下标从1开始
             if (a[i] +  < n)//当a[i] = n，不操作
                 ans += query(a[i] + , n, , n, );
         }
         minans = ans;
         for (i = ; i < n; i++)
         {
             ans += n - a[i] *  -;
             if (minans > ans)
                 minans = ans;
         }
         printf("%d\n", minans);
     }
     return ;
 }