Minimum Inversion Number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15247 Accepted Submission(s): 9311
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
Sample Output
16
Author
CHEN, Gaoli
Source
Recommend
虽然在学线段树。。。但是感觉美必要为了用线段树而用线段树。。。
我知道树状数组的东西都可以用线段树来解决。。。。
不过还是觉得树状数组。。。实在是优美。。。所以这题还是用树状数组搞得。。
这题是问一个长度为n的循环数组中,逆序对最少的个数。。。
我们可以先用树状数组求出初始的数列的逆序对。。。
然后其他的可以通过递推得到。。。。
当a[i]从处于位置1而被放到最后的时候。。。
cnt = cnt -a[i]+n-a[i]+1;
然后取所有cnt的最大值就行。
1 /************************************************************************* 2 > File Name: code/hud/1394.cpp 3 > Author: 111qqz 4 > Email: rkz2013@126.com 5 > Created Time: 2015年10月28日 星期三 21时16分47秒 6 ************************************************************************/ 7 8 #include<iostream> 9 #include<iomanip> 10 #include<cstdio> 11 #include<algorithm> 12 #include<cmath> 13 #include<cstring> 14 #include<string> 15 #include<map> 16 #include<set> 17 #include<queue> 18 #include<vector> 19 #include<stack> 20 #include<cctype> 21 22 #define yn hez111qqz 23 #define j1 cute111qqz 24 #define ms(a,x) memset(a,x,sizeof(a)) 25 using namespace std; 26 const int dx4[4]={1,0,0,-1}; 27 const int dy4[4]={0,-1,1,0}; 28 typedef long long LL; 29 typedef double DB; 30 const int inf = 0x3f3f3f3f; 31 const int N=5E3+7; 32 int c[N]; 33 int n,a[N]; 34 35 int lowbit( int x) 36 { 37 return x&(-x); 38 } 39 40 41 void update ( int x,int delta) 42 { 43 // cout<<"a?"<<endl; 44 for ( int i = x ; i <= n ; i = i + lowbit(i)) c[i] = c[i] + delta; 45 } 46 int Sum( int x) 47 { 48 // cout<<"b?"<<endl; 49 int res = 0 ; 50 for ( int i = x ;i >= 1; i = i - lowbit(i)) 51 { 52 res = res + c[i]; 53 } 54 return res; 55 } 56 int main() 57 { 58 #ifndef ONLINE_JUDGE 59 freopen("in.txt","r",stdin); 60 #endif 61 while (scanf("%d",&n)!=EOF) 62 { 63 ms(c,0); 64 for ( int i = 1 ; i <= n ; i++) 65 { 66 scanf("%d",&a[i]); 67 a[i]++; 68 } 69 int cnt = 0 ; 70 int ans = inf; 71 // puts("mmmmmmmiao?"); 72 for ( int i = 1 ; i <= n ; i++) 73 { 74 update(a[i],1); 75 cnt = cnt + i - Sum(a[i]); 76 // puts("whats wrong?"); 77 } 78 // printf("cnt::%d ",cnt); 79 if (cnt<ans) ans = cnt; 80 for ( int i = 1 ; i <= n ; i++) 81 { 82 cnt =cnt -a[i]+n-a[i]+1; 83 if (cnt<ans&&cnt>0) ans = cnt; 84 // printf("cnt:%d ",cnt); 85 } 86 printf("%d ",ans); 87 } 88 89 90 91 #ifndef ONLINE_JUDGE 92 fclose(stdin); 93 #endif 94 return 0; 95 }