HDU 1394 Minimum Inversion Number

分手后的思念是犯贱 2022-09-20 15:30 160阅读 0赞

# Minimum Inversion Number #

**Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)  
Total Submission(s): 10207 Accepted Submission(s): 6274**

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][ZOJ Monthly_ January 2003]

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Ignatius.L | We have carefully selected several similar problems for you: [1166][] [1698][] [1540][] [1542][] [1255][]

/*
    	开始看半天，不知咋做，其实看清楚题意后很简单，首先，有n个数，并且这n个数还是从0~n-1的。这条件很总要。
    因为 从这可以知道， 数组的第一个元素a[0]，后面比他小的数有a[0]个，若把a[0]放到末尾，则a[0]前面比他大的
    数有(n-1)-a[0]个！！！
    可以不用线段树做的。。。。。。
    */
    #include <stdio.h>
    struct node
    {
    	int l, r;
    	int sum;
    }tree[20000];
    int h[5005];
    int sum;
    
    int MIN(int a, int b)
    {
    	if(a<b)
    		return a;
    	return b;
    }
    void build(int l, int r, int n)
    {
    	int mid;
    	tree[n].l = l, tree[n].r = r;
    	tree[n].sum = 0;		//建树初始化
    	if(l==r)
    		return ;
    	mid = (tree[n].l + tree[n].r)/2;
    
    	build(l, mid, 2*n);
    	build(mid+1, r, 2*n+1);
    }
    void add(int l, int r, int n)
    {
    	int mid;
    
    	tree[n].sum += 1;		//把包含h[i]的区间都加1，代表此区间有多少个元素
    	if(tree[n].l==tree[n].r)
    		return ;
    	mid = (tree[n].l+tree[n].r)>>1;
    	if(l>=mid+1)
    		add(l, r, 2*n+1);
    	else if(r<=mid)
    		add(l, r, 2*n);
    	else
    	{
    		add(l, mid, 2*n);
    		add(mid+1, r, 2*n+1);
    	}
    }
    void qu(int l, int r, int n)
    {
    	int mid;
    	if(tree[n].l==l && tree[n].r==r)
    	{
    		sum += tree[n].sum;
    		return;
    	}
    	mid = (tree[n].l+tree[n].r)>>1;
    	if(l>=mid+1)
    		qu(l, r, 2*n+1);
    	else if(r<=mid)
    		qu(l, r, 2*n);
    	else
    	{
    		qu(l, mid, 2*n);
    		qu(mid+1, r, 2*n+1);
    	}
    }
    int main()
    {
    	int n;
    	int i, k;
    
    	while(scanf("%d", &n) != -1)
    	{
    		build(0, n-1, 1);
    		
    		sum = 0;
    		for(i=0; i<n; i++)
    		{
    			scanf("%d", &h[i]);
    			qu(h[i], n-1, 1);		//查找从 h[i]~n-1 间有多少个元素，即有多少逆序数对
    			add(h[i],h[i], 1);			
    		}
    		k = sum;
    		for(i=0; i<n-1; i++)
    		{
    			k += (n-1)-h[i];	//把数组循环移位，增加的逆序数对
    			k -= h[i];			//减少的逆序数对
    			sum = MIN(k, sum);
    		}
    		printf("%d\n", sum);
    	}
    	return 0;
    }

[ZOJ Monthly_ January 2003]: http://acm.hdu.edu.cn/search.php?field=problem&key=ZOJ%20Monthly,%20January%202003&source=1&searchmode=source
[1166]: http://acm.hdu.edu.cn/showproblem.php?pid=1166
[1698]: http://acm.hdu.edu.cn/showproblem.php?pid=1698
[1540]: http://acm.hdu.edu.cn/showproblem.php?pid=1540
[1542]: http://acm.hdu.edu.cn/showproblem.php?pid=1542
[1255]: http://acm.hdu.edu.cn/showproblem.php?pid=1255