HDU 1394 Minimum Inversion Number【多数列逆序对】

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本文探讨了在给定序列中寻找最小逆序对数量的算法。通过移动序列的前m个元素到序列末尾，生成多种可能的序列排列，并采用树状数组和递推方法高效计算每种排列的逆序数，最终确定逆序数最小的排列。

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 18906 Accepted Submission(s): 11415

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

Author

CHEN, Gaoli

Source

ZOJ Monthly, January 2003

Recommend

Ignatius.L

刚开始思路错了，逆序对用树状数组只能动态求！！！

利用数组的值为0-N-1这个特点就可以用递推求出来了；

AC代码：

#include<cstdio>
#include<algorithm>

using namespace std;

struct Node{
	int val,id;
}a[10011];
int C[10011],f[5055];

bool cmp(Node a,Node b) {
	if(a.val!=b.val) return a.val>b.val;
	return a.id>b.id;
}

int Lowbit(int x) {
	return x&(-x);
}
void Add(int x,int d,int N) {
	while(x<=N) {
		C[x]+=d; x+=Lowbit(x);
	}
}
int Sum(int x) {
	int ret=0;
	while(x>0) {
		ret+=C[x]; x-=Lowbit(x);
	}
	return ret;
}

int main() {
	int N; 
	while(~scanf("%d",&N)) {
		for(int i=0;i<N;++i) {
			scanf("%d",&a[i].val); 
			a[i].id=i+1;  
		}
		sort(a,a+N,cmp);
		for(int i=0;i<N;++i) f[a[i].id-1]=i;
	    for(int i=0;i<=N;++i) C[i]=0;
	    int ans=0;
		for(int i=0;i<N;++i) {
			Add(a[i].id,1,N); 
			ans=ans+Sum(a[i].id-1); 
		}
		int pre=ans;// printf("%d\n",ans);
		for(int i=0;i<N-1;++i) {
			pre=pre-(a[f[i]].val)+(N-1-a[f[i]].val);
		//	printf("%d\n",pre);
			ans=min(ans,pre);
		}
		printf("%d\n",ans);
	}
	return 0;
}