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  • [hdu1394]Minimum Inversion Number(树状数组)

    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
    10 1 3 6 9 0 8 5 7 4 2
     
    Sample Output
    16
     
    Author
    CHEN, Gaoli
     
    Source
     
    Recommend
    Ignatius.L
     
     
    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
    ",mn);
    44     }
    45     
    46     return 0;
    47 }
     
     
     
     
     
     
     
     
     
     
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  • 原文地址:https://www.cnblogs.com/Pumbit-Legion/p/5874029.html
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