HDU 1394 Minimum Inversion Number

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C - Minimum Inversion Number
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
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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

 

題目大意為 有n個從0到n 的數 按照給定次序排列 每次只能將最前的一個數移到末尾 算出每種情況的逆序數 並求出逆序數最小值

 

這道題屬於 歸併排序的應用

首先 我們把當前情況下的逆序數求出

然後迴圈n次 每次最前一個往最後移 則 ans=ans+ (n-1+a[i])-a[i]

#include<stdio.h>#include<string.h>#include<math.h>#include<iostream>#include<algorithm>#include<queue>#include<stack>#define mem(a,b) memset(a,b,sizeof(a))#define ll __int64#define MAXN 1000#define INF 0x7ffffff#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1using namespace std;int num[ 5000+10],a[5000+10];int vis[5000+10];int ans;void Merge(int begin,int mid,int end){    int i=begin,j=mid+1,k=begin;    while(i<=mid&&j<=end)    {        if(a[i]<=a[j])                    vis[k++]=a[i++];                else        {            ans+=mid-i+1;            vis[k++]=a[j++];        }    }    while(i<=mid)    {        vis[k++]=a[i++];    }    while(j<=end)    {        vis[k++]=a[j++];    }    for(i=begin;i<=end;i++)    {        a[i]=vis[i];    }}void mergesort(int begin,int end){   if(begin!=end)   {       int mid=(begin+end)/2;       mergesort(begin,mid);       mergesort(mid+1,end);       Merge(begin,mid,end);   }}int main(){    int n,i,j,minn;    while(scanf("%d",&n)!=EOF)    {      ans=0;      for(i=1;i<=n;i++)        {scanf("%d",&a[i]); num[i]=a[i];}      mergesort(1,n);      minn=ans;      for(i=1;i<n;i++)      {          ans=ans+n-1-2*num[i];          if(ans<minn) minn=ans;      }      cout<<minn<<endl;    }    return 0;}