HDU 1394 Minimum Inversion Number (线段树&&暴力)

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10477 Accepted Submission(s): 6464


[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
For each case, output the minimum inversion number on a single line.

[align=left]Sample Input[/align]

10
1 3 6 9 0 8 5 7 4 2

[align=left]Sample Output[/align]

16

[align=left]Author[/align]
CHEN, Gaoli

[align=left]Source[/align]
ZOJ Monthly, January 2003

[align=left]Recommend[/align]
Ignatius.L

题目的意思就是说给你一个序列，求这个序列的逆序数的对数，然后将第一个元素放到最后，从而生成了一个新的序列，直到最后一个元素到第一个时停止，求生成的序列中，最少的逆序数的对数是多少。

我谈谈我的做法
1.暴力，每次找出新序列的逆序数的对数，然后看看是不是最小值，最后输出最小值即可
但这里要点技巧，在寻找新的序列的逆序数的对数时，没有必要去扫一遍出结果。
可以想想，因为序列的数字是从0~n-1连续的，那么当第一个数移到最后去的时候，原来比这个数大的数字变成了逆序数，比这个数小的数就不是逆序数了
举个例子：
3 2 4 5
本来 3 2是逆序数，3 4,3 5不是逆序数，当序列变成2 4 5 3时，原来的3 4变成了4 3是逆序数，2 3就不是逆序数了

所以在原有的序列中比第一个元素要小的数个数为(即逆序数的对数) low[a[i]]=a[i]，所以可以推出比第一个元素要大的数的个数为up[a[i]]=n-1-a[i]
那么新的序列中逆序数的对数是 sum=sum-low[a[i]]+up[a[i]]=sum-a[i]+(n-1-a[i])

暴力的代码

#include<cstdio>
#include<cstring>
#include<stdlib.h>
#include<algorithm>
using namespace std;
struct node
{
int l,r;
int num;
int mid()
{
return (l+r)>>1;
}
}a[5000*4];
int b[5005];
void btree(int l,int r,int step)
{
a[step].l=l;
a[step].r=r;
a[step].num=0;
if(l==r)  return ;
int mid=a[step].mid();
btree(l,mid,step*2);
btree(mid+1,r,step*2+1);
}

void ptree(int step,int vis)
{
a[step].num++;
if(a[step].l==a[step].r)  return ;
int mid=a[step].mid();
if(vis>mid)
ptree(step*2+1,vis);
else
ptree(step*2,vis);
}

int fintree(int step,int x,int y)
{
if(a[step].l==x&&a[step].r==y)
return a[step].num;
int mid=a[step].mid();
if(x>mid)
return fintree(step*2+1,x,y);
else if(y<=mid)
return fintree(step*2,x,y);
else
return fintree(step*2,x,mid)+fintree(step*2+1,mid+1,y);
}
int main()
{
int n,i,j;
while(scanf("%d",&n)!=EOF)
{
int ans=0;
btree(0,n-1,1);
for(i=0;i<n;i++)
{
scanf("%d",&b[i]);
ans+=fintree(1,b[i],n-1);
ptree(1,b[i]);
}
int minn=ans;
for(i=0;i<n;i++)
{
ans=ans-b[i]+(n-1-b[i]);
if(minn>ans) minn=ans;
}
printf("%d\n",minn);
}
return 0;
}

View Code