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HDOJ 1394 Minimum Inversion Number 求循环串的最小逆序数(暴

来源:互联网 作者:佚名 时间:2015-09-01 23:54
Java版本说明: Java开发中,经常遇到从GBK到Unicode码的转换;大家经常做的做法,大概有2种,一种是:native2ascii;只要保存好文件,每次编辑好文件,然后使

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

10 1 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:a using 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!

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