HDOJ 1394 Minimum Inversion Number（线段树+逆序数）

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 21463    Accepted Submission(s): 12851

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

101 3 6 9 0 8 5 7 4 2

 

Sample Output

16

 

Author

CHEN, Gaoli

 

Source

ZOJ Monthly, January 2003

 

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建立一个空的线段树，每加入一个元素，查询线段树里大于这个元素的的个数有多少个，即此时产生了多少对逆序数。

当建立好线段树以后，每将头位置的数例如k，放到末尾。减少了k个逆序数，因为在变换之前，K位置后面，有K个数小于K；同时又增加了n-k-1个逆序数，因为变换之后，k前面有n-1-k个数大于k.

如: 2 1 0   逆序数 3

      1 0 2   逆序数 1（3-2+2-2）

      0 2 1   逆序数 1 （1-1+2-1）

#include<iostream>#include<algorithm>#include<cstdio>#include<cstring>#include<cmath>#include<string>const int MAX_N = 5000;int pos;int dat[MAX_N<<2];int a[MAX_N + 10];using namespace std;void update(int k)//更新{while (k>0){k = (k - 1) / 2;dat[k] = dat[2 * k + 1] + dat[2 * k + 2];}}//求[a,b)之间的和//k 是节点编号， l和r表示对应[l,r)区间int querya(int a, int b,int k,int l,int r){int ans=0;//[a,b)和[l,r)不相交if (r <= a || b <= l)return 0;//[a,b)完全包含[l,r)if (a <= l&&r <= b)return dat[k];else {//左边的和+右边的和ans += querya(a, b, k * 2 + 1, l, (l + r) / 2);ans += querya(a, b, k * 2 + 2, (l + r) / 2,r );return ans;}}int main(){int n;while (scanf("%d",&n)!=EOF){memset(dat, 0, sizeof(dat));int sum = 0;pos = 1;while (pos<n){pos *= 2;}pos--;//叶子的第一个节点是pos-1for (int i = 0; i < n; i++){scanf("%d", &a[i]);int t = a[i];dat[pos + t] = 1;update(pos + t);sum += querya(0, pos + 1, 0, 0, pos + 1) - querya(0, t + 1, 0, 0, pos + 1);}int ans = sum;for (int i = 0; i < n; i++){sum = sum - a[i] + n - 1 - a[i];ans = min(sum, ans);}printf("%d\n", ans);}return 0;}