HDOJ 1394 Minimum Inversion Number 求循环串的最小逆序数（暴力&&线段树）

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14879    Accepted Submission(s): 9082

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

ac代码：

暴力解法：425ms

#include<stdio.h>#include<string.h>#include<math.h>#include<iostream>#include<algorithm>#define INF 0x7fffffff#define MAXN 10010#define max(a,b) a>b?a:b#define min(a,b) a>b?b:ausing namespace std;int num[MAXN];int main(){int i,j,n;while(scanf("%d",&n)!=EOF){int M;int cnt=0;for(i=0;i<n;i++)scanf("%d",&num[i]);for(i=0;i<n;i++){for(j=i+1;j<n;j++){if(num[j]<num[i])cnt++;}}M=cnt;for(i=0;i<n;i++){cnt=cnt-num[i]+n-1-num[i];if(M>cnt)M=cnt;}printf("%d\n",M);}return 0;}

看到很多人都用线段树来写，我也想写一下，明天吧！