标签：des blog os io for ar div line
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
题目大意为 有n个从0到n 的数 按照给定次序排列 每次只能将最前的一个数移到末尾 算出每种情况的逆序数 并求出逆序数最小值
这道题属于 归并排序的应用
首先 我们把当前情况下的逆序数求出
然后循环n次 每次最前一个往最后移 则 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|1
using 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;
}
HDU 1394 Minimum Inversion Number,布布扣,bubuko.com
HDU 1394 Minimum Inversion Number
标签：des blog os io for ar div line
原文地址：http://www.cnblogs.com/sola1994/p/3901536.html
