hdu 1394 Minimum Inversion Number(这道题改日我要用线段树再做一次哟~)
2016-03-29 23:23
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[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
//有的大神是用线段树解的,原谅我怎么都无法看出线段树来。。。。_(:з」∠)_
//于是采用暴力解,然而。。超时了 (╯‵□′)╯︵┴─┴
//于是借鉴别人的代码~~~ (●'◡'●)
//思路:每次把末尾的数掉到序列前面时,减少的逆序对数为n-1-a[i] ,
//增加的逆序对数为a[i] ,这样就可在所有的序列中找出含有逆序对最少的了!(ps:给跪了,这是性质吗。。。
//再贴上我的狗血超时代码。。。
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
//有的大神是用线段树解的,原谅我怎么都无法看出线段树来。。。。_(:з」∠)_
//于是采用暴力解,然而。。超时了 (╯‵□′)╯︵┴─┴
//于是借鉴别人的代码~~~ (●'◡'●)
//思路:每次把末尾的数掉到序列前面时,减少的逆序对数为n-1-a[i] ,
//增加的逆序对数为a[i] ,这样就可在所有的序列中找出含有逆序对最少的了!(ps:给跪了,这是性质吗。。。
#include <iostream> #include <cstdio> using namespace std; int main() { int n,ans,k; int data[5005]; while(cin>>n) { ans=0; for(int i=0;i<n;i++) scanf("%d",&data[i]); for(int i=0;i<n;i++) { for(int j=i;j<n;j++) { if(data[i]>data[j]) ans++; } } k=ans; for(int i=n-1;i>=0;i--) { k-=n-1-data[i]; k+=data[i]; if(ans>k) ans=k; } cout<<ans<<endl; } return 0; }
//再贴上我的狗血超时代码。。。
#include <iostream> #include <cstdio> #include <algorithm> using namespace std; int find(int a[],int n) { int ans=0; for(int i=0;i<n;i++) { for(int j=i+1;j<n;j++) { if(a[i]>a[j]) ans++; } } return ans; } int main() { int n; int data[5005],num[5005]; while(scanf("%d",&n)!=-1) { for(int i=0;i<n;i++) scanf("%d",&data[i]); for(int i=0;i<n;i++) { num[i]=find(data,n); int tmp=data[n-1]; for(int j=n-1;j>=1;j--) { data[j]=data[j-1]; } data[0]=tmp; } sort(num,num+n); cout<<num[0]<<endl; } return 0; }
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