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[hdu1394]Minimum Inversion Number(树状数组)

2024-08-13 10:10:24   224人阅读

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18395    Accepted Submission(s): 11168


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.
 

 

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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更更更
 
隔壁YNY看到就直接暴力，当然T了(哈哈哈一起笑他)
动动脑子，题目说是一个环
那么每转一次，就相当于把第一个数放到最后面
考虑第一个数对原有答案的贡献是a[1]-1，也就是小于它的个数(数据是1到n的排列)
最后一个数对原答案的贡献相反
那么移动后当前逆序对数就要减去比第一个数小的个数，再加上比它大的数的个数
这样我们求出一次移动后的逆序对数
这时候我们发现下一次移动直接修改答案就好，不需要改动树状数组
推出式子ans+=n-a[i]-(a[i]-1)
 1 #include<stdio.h> 2 #include<stdlib.h> 3 #include<string.h> 4 #define LL long long 5 int bit[5050]={0},n,a[5050]; 6 inline LL min(LL a,LL b){ 7     return a<b?a:b; 8 } 9 inline int lb(int x){10     return x&(-x);11 }12 inline LL q(int x){13     LL ans=0;14     while(x){15         ans+=bit[x];16         x-=lb(x);17     }18     return ans;19 }20 inline int c(int x){21     while(x<=n){22         bit[x]++;23         x+=lb(x);24     }25     return 0;26 }27 int main(){28     while(scanf("%d",&n)!=EOF){29         memset(bit,0,sizeof(bit));30         LL ans=0;31         for(int i=1;i<=n;i++){32             scanf("%d",&a[i]);33             a[i]++;34             ans+=q(n)-q(a[i]);35             c(a[i]);36         }37         LL mn=ans;38         mn=min(mn,ans);39         for(int i=1;i<=n;i++){40             ans+=n-a[i]-(a[i]-1);41             mn=min(mn,ans);42         }43         printf("%lld\n",mn);44     }45     46     return 0;47 }

 

 
 
 
 
 
 
 
 
 
 

[hdu1394]Minimum Inversion Number(树状数组)


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