Minimum Inversion Number hdu 1394
Minimum Inversion Number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1990 Accepted Submission(s): 1188
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.
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
1 3 6 9 0 8 5 7 4 2
Sample Output
16
Author
CHEN, Gaoli
Source
1 //此题可推出公式,当把队首元素P放到后面去时,逆序数和变化为 N - 1 - P - N
2 #include <stdio.h>
3 #include <string.h>
4 #include <stdlib.h>
5 #include <algorithm>
6
7 using namespace std;
8
9 int dp[5010];
10 int sum;
11
12 void fun(int N)
13 {
14 int i, j;
15 sum = 0;
16
17 for(i = 0; i < N; i++)
18 for(j = i + 1; j < N; j++)
19 if (dp[i] > dp[j])
20 sum++;
21 }
22
23 int move(int N )
24 {
25 int i, j, p;
26 p = sum;
27
28 for(i = 0; i < N; i++)
29 {
30
31 p = p - dp[i] + N - 1 - dp[i];
32 //printf("%d\n",p);
33 if (p < sum)
34 sum = p;
35 }
36
37 return sum;
38
39 }
40
41 int main( )
42 {
43 int N, i, j;
44 while(scanf("%d",&N) != EOF)
45 {
46 for(i = 0; i < N; i++)
47 scanf("%d",&dp[i]);
48 fun(N);
49 printf("%d\n", move(N));
50 }
51 return 0;
52 }
posted on 2011-08-17 00:41 more think, more gains 阅读(161) 评论(0) 编辑 收藏 举报