hdu 1394 Minimum Inversion Number(樹狀數組)

Minimum Inversion Number

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 26   Accepted Submission(s) : 20

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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

該題目的意思  看了好久 ，最後參考博客才明白，就是把一個序列，

第一個 到n-1個 依次移到最後，全部移完之後，看有多少的最少的

逆序隊。  解題思路大致這樣， 累加該處到m的逆序對數目，然後，再

依次將1-n-1的數移到末尾。 根據減少的逆序對爲 a【0】，增加的爲m-a【0】+1；

最後 比較  求得最小的逆序對數目；

#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<string.h>
#define maxn 5050
using namespace std;
int ss[maxn]; //定義 保存書 
int a[maxn];
int lowbit(int x)
{
	return x&(-x);
}

int sum(int x)  //求和 
{
	int suM=0;
	while(x>0)
	{
		suM+=ss[x];
		x-=lowbit(x);
	}
	return suM;
}

void updata(int x,int y) //加減 
{
	while(x<maxn)
	{
		ss[x]+=y;
		x+=lowbit(x);
	}
}
int main()
{
	int m,b,c;
	while(~scanf("%d",&m))
	{
		memset(ss,0,sizeof(ss));
		int ans=0;
		for(int i=1;i<=m;i++)
		{
			scanf("%d",&a[i]);
			a[i]++;
			ans+=sum(m)-sum(a[i]);
			updata(a[i],1);
		}
		int min=ans;
		for(int i=1;i<=m;i++)
		{
			ans+=m-a[i]-(a[i]-1);
			if(ans<min) min=ans;
		}
		printf("%d\n",min);
	}
	return 0;
}

而本題也是好用 線段樹來寫的，先建立一棵線段樹，然後輸入a數組的值，

順求得逆序數ans的初值，然後更新節點區域的值，注意子節點樹之和爲

父節點數，最後，，因爲每次移動減少的是a【i】個逆序對，同時增加了

n-a【i】-1，，具體代碼實現如下：

#include<stdio.h>
#include<math.h>
struct shu
{
	int left;
	int right;
	int num;
}s[15010];
int n;
int a[5010];


int min(int a,int b)
{
	return a>b?b:a;
}

void build(int ll,int rr,int i)  //建立一棵線段樹 
{
	s[i].right =rr;
	s[i].left =ll;
	s[i].num =0;
	if(s[i].left==s[i].right)
	{
		return;
	}
	int mm;
	mm=(s[i].left +s[i].right )/2;
	build(ll,mm,2*i);
	build(mm+1,rr,2*i+1);
}

int sum(int ll,int rr,int i) //求線段域的值 
{
	int ans=0;
	if(ll==s[i].left &&rr==s[i].right )
	return s[i].num;
	int mm;
	mm=(s[i].left+s[i].right)/2;
	if(rr<=mm)
	ans+=sum(ll,rr,2*i);
	else if(ll>mm) 
	ans+=sum(ll,rr,2*i+1);
	else 
	{
		ans+=sum(ll,mm,2*i);
		ans+=sum(mm+1,rr,2*i+1);
	}
	return ans;
}


void updata(int xx,int i) // 
{
	if(s[i].left ==s[i].right)
	{
		s[i].num ++;
		return;
	}
	int mm=(s[i].left +s[i].right)/2;
	if(xx<=mm) updata(xx,2*i);
	else if(xx>=mm+1)  updata(xx,2*i+1);
	s[i].num=s[2*i].num +s[2*i+1].num;  //父節點爲兩個節點的和 
}


int main()
{
	int i;
	while(~scanf("%d",&n))
	{
		build(0,n-1,1);
		int ans=0;
		for(i=0;i<n;i++)
		{
			scanf("%d",&a[i]);
			ans+=sum(a[i],n-1,1); //初次求得逆序對 
			updata(a[i],1);
		}
		
		int tt=ans;
		for(i=0;i<n;i++)
		{
			tt=tt+n-1-a[i]-a[i]; //移動引起的節點的增加減少 
			ans=min(ans,tt); //比較，求小的值 
		}
		printf("%d\n",ans);
	}
	return 0;
}