hdu 1394 逆序对

Minimum Inversion Number

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

Total Submission(s): 4067 Accepted Submission(s): 2446



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

algorithm：
1：求出序列的逆序对数ct（归并排序||树状数组||线段树）
2：按题扫一遍，ct += n - 2*a[i] -1;

归并：

#include <iostream>
#include <cstring>
#define INF 0x3f3f3f3f
#define M 5010
#define clr(x,k) memset((x),(k),sizeof(x))
using namespace std;
int a[M], b[M], c[M], n, ct;
void merge(int l, int mid, int r)
{
	int i = 0, j = l, k = mid +1;
	while (j<=mid && k<=r)
	{
		if (a[j]>a[k])
		{
			//p (a[k], a[j]);
			b[i++] = a[k++];
			ct += (mid - j + 1);
		}
		else
		{
			b[i++] = a[j++];
		}
	}
	while (j<=mid) b[i++] = a[j++];
	while (k<=r)   b[i++] = a[k++];
	for (j=0; j<i;j++)
	a[l+j] = b[j];
}

void mergesort(int l, int r)
{
	if (l<r)
	{
		int mid = (l+r)/2;
		mergesort(l, mid);
		mergesort(mid+1, r);
		merge(l, mid, r);
	}
	return ;
}

int main()
{
	while (scanf("%d",&n)!=EOF)
	{
    ct = 0;
		clr(a,0);
		clr(b,0);
		for (int i=0;i<n;i++)
		{
			scanf("%d",&a[i]);
			c[i] = a[i];
		}
		mergesort(0,n-1);
		int mmm = ct;
		for (int i=0;i<n-1;i++)
		{
        		ct += n - 2*c[i] -1;
			if (mmm>ct) mmm = ct;
		}
		printf("%d\n",mmm);
	}
	return 0;
}

树状数组：

#include <iostream>
#include <cstring>
#define INF 0x3f3f3f3f
#define M 5010
#define clr(x,k) memset((x),(k),sizeof(x))
#define lowbit(w) (w & (-w))
using namespace std;

int a[M], b[M], c[M];
int n, m, sum;

void modify(int x)
{
	while (x<M)
	{
		c[x]++;
		x += lowbit(x);
	}
}

int Sum(int x)
{
	int sum = 0;
	while (x>0)
	{
		sum += c[x];
		x -= lowbit(x);
	}
	return sum;
}

int main()
{
	while (scanf("%d", &n)!=EOF)
	{
		clr(c, 0);
		clr(b, 0);
		for (int i=0;i<n;++i)
		{
			scanf("%d",&a[i]);
			b[i] = Sum(a[i]+1);
			modify(a[i]+1);
		}
		int ct = 0;
		for (int i=0;i<n;i++)
		{
			ct += a[i]-b[i];
		}
		int mmm = ct;
		for (int i=0;i<n-1;i++)
		{
        		ct += n - 2*a[i] -1;
			if (mmm>ct) mmm = ct;
		}
		printf("%d\n",mmm);
	}
	return 0;
}

线段树：

#include<cstdio>
#include<algorithm>
using namespace std;
int sum[5555<<2];
int query(int x,int y,int s,int t,int r)
{
	if((x<=s)&&(y>=t))
		return sum[r];
	int m=(s+t)>>1;
	int ans=0;
	if(x<=m) ans+=query(x,y,s,m,r<<1);
	if(y>m) ans+=query(x,y,m+1,t,r<<1|1);
	return ans;
}
void update(int p,int s,int t,int r)
{
	if(s==t)
	{
		sum[r]++;
		return ;
	}
	int m=(s+t)>>1;
	if(p<=m) update(p,s,m,r<<1);
	else update(p,m+1,t,r<<1|1);
	sum[r]=sum[r<<1]+sum[r<<1|1];
}
int main()
{
	int n;
	while(~scanf("%d",&n))
	{
		int ans=0;
		int a[5555];
		memset(sum,0,sizeof(sum));
		for(int i=0;i<n;i++)
		{
			scanf("%d",&a[i]);
			ans+=query(a[i]+1,n-1,0,n-1,1);		
			update(a[i],0,n-1,1);
		}
		int minn=ans;
		for(int i=0;i<n-1;i++)
		{
			ans+=n-a[i]-a[i]-1;
			if(ans<minn) minn=ans;
		}
		printf("%d\n",minn);
	}	
	return 0;
}

ps：线段树不是我写的。。风格。。。