hdu 1394 解题报告 Minimum Inversion Number
Minimum Inversion Number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6820 Accepted Submission(s): 4162
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.
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
又是一个经典的线段树，首先要知道如何求一个序列的逆序——
比如2514 ：
2之前没有数大于它，所以为0，
5之前也没有数大于它，所以为0， 1之前2,5都大于它，所以为2， 4之前只有5大于它，所以为1，
因此2514的逆序数为：0 + 0 + 2 + 1 = 3; 从前面的描述中，我们可以发现，只要依次累计前面有几个数大于当前数即可。于是我们可以用一个数组l[n]（初始化为0），然后每读取一个数x[i] 就标识l[x[i]] += 1 , 但是在标识之前我们先要统计有多少个数大于x[i],即累计l[x[i]+1] + ... +l[n-1]，很显然累计的时间复杂度为O(N)，我们能不能在快点呢?
那么如何求最小逆序数呢？（我这里假设一个序列中每个数字都不同） 若abcde...的逆序数为sum，那么bcde...a的逆序数是多少？我们假设abcde...中小于a的个数为x[i] , 那么大于a的个数就是n - x[i]，当把a移动左移一位时，原来比a大的现在都成了a的逆序对，故减少了x[i]+1，即 相比而言总共加了n-x[i]-x[i]-1;其实，线段树只有开始用了一次，后面求最小值就没用了，所以用暴力应该也能过，sum再更新N—－1次就可以得到最小值了，不算复杂！
于是我
#include <iostream>
#include<stdio.h>
using namespace std;
#define N 50000
int l[N<<2],x[N];
void build(int num ,int s,int e)
{
l[num]=0;
if(s==e)
return ;
int mid=(s+e)>>1;
build(num<<1,s,mid);
build(num<<1|1,mid+1,e);
}
int query(int num ,int s,int e ,int a,int b)
{
if(a<=s&&b>=e)
return l[num];
int mid=(s+e)>>1;
int ret=0;
if(mid>=a)
ret+=query(num<<1,s,mid,a,b);
if(mid<b)
ret+=query(num<<1|1,mid+1,e,a,b);
return ret;
}
void update(int num ,int s,int e,int a)
{
if(s==e)
{
l[num]++;
return ;
}
int mid=(s+e)>>1;
if(a<=mid)
update(num<<1,s,mid,a);
else
update(num<<1|1,mid+1,e,a);
l[num]=l[num<<1]+l[num<<1|1];
}
int main()
{
int n,i;
while(scanf("%d",&n)!=EOF)
{
build(1,0,n-1);
int sum=0;
for(i=0;i<n;i++)
{
scanf("%d",&x[i]);
sum=sum+query(1,0,n-1,x[i],n-1);
update(1,0,n-1,x[i]);
}
int re=sum;
for(i=0;i<n-1;i++)
{
sum+=n-x[i]-x[i]-1;
if(sum<re)
re=sum;
}
printf("%d\n",re);
}
return 0;
}