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HDU 1394 Minimum Inversion Number
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
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
101 3 6 9 0 8 5 7 4 2
Sample Output
16
Author
CHEN, Gaoli
Source
ZOJ Monthly, January 2003
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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
1 #include<cstdio> 2 #include<cstring> 3 #include<stdlib.h> 4 #include<algorithm> 5 using namespace std; 6 int a[5005]; 7 int main() 8 { 9 int n,i,j;10 while(scanf("%d",&n)!=EOF)11 {12 int sum=0;13 for(i=0;i<n;i++)14 {15 scanf("%d",&a[i]);16 for(j=0;j<i;j++)17 if(a[j]>a[i])18 sum++;19 }20 int ans=sum;21 for(i=0;i<n;i++)22 {23 sum=sum-a[i]+(n-a[i]-1);24 if(sum<ans)25 ans=sum;26 }27 printf("%d\n",ans);28 }29 return 0;30 }
2.线段树
同样的一开始也是要找出逆序数的对数，暴力方法是从第一个数到现在这个数扫一遍查找，线段树的方法就是用线段树去查询，会更快
后面对新序列的逆序数对数的处理时相同的
线段树的代码
1 #include<cstdio> 2 #include<cstring> 3 #include<stdlib.h> 4 #include<algorithm> 5 using namespace std; 6 struct node 7 { 8 int l,r; 9 int num;10 int mid()11 {12 return (l+r)>>1;13 }14 }a[5000*4];15 int b[5005];16 void btree(int l,int r,int step)17 {18 a[step].l=l;19 a[step].r=r;20 a[step].num=0;21 if(l==r) return ;22 int mid=a[step].mid();23 btree(l,mid,step*2);24 btree(mid+1,r,step*2+1);25 }26 27 void ptree(int step,int vis)28 {29 a[step].num++;30 if(a[step].l==a[step].r) return ;31 int mid=a[step].mid();32 if(vis>mid)33 ptree(step*2+1,vis);34 else35 ptree(step*2,vis);36 }37 38 int fintree(int step,int x,int y)39 {40 if(a[step].l==x&&a[step].r==y)41 return a[step].num;42 int mid=a[step].mid();43 if(x>mid)44 return fintree(step*2+1,x,y);45 else if(y<=mid)46 return fintree(step*2,x,y);47 else48 return fintree(step*2,x,mid)+fintree(step*2+1,mid+1,y);49 }50 int main()51 {52 int n,i,j;53 while(scanf("%d",&n)!=EOF)54 {55 int ans=0;56 btree(0,n-1,1);57 for(i=0;i<n;i++)58 {59 scanf("%d",&b[i]);60 ans+=fintree(1,b[i],n-1);61 ptree(1,b[i]);62 }63 int minn=ans;64 for(i=0;i<n;i++)65 {66 ans=ans-b[i]+(n-1-b[i]);67 if(minn>ans) minn=ans;68 }69 printf("%d\n",minn);70 }71 return 0;72 }
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