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Minimum Inversion Number(线段树求逆序数)

热度:26   发布时间:2024-01-13 21:20:58.0

点击打开链接Minimum Inversion Number

有一点没想明白,今天突然间想明白了,sum[rt]记录的是rt表示的区间有几个数,而输入一个数,查询一次,查询的目的就是返回当前区间内已经输入了多少个数,查询区间是(x,n-1)是因为要找总共有多少个逆序数,比如说先输入5,再输入3,那么输入3时查询前边输入的数有几个比三大的即当前这些数的总的逆序数,因为输入3,前边找到一个比3大的数,那么就看5,3,可以知道,5由于3的输入,逆序数加1,所以有用

ans += query(x[i] , n - 1 , 0 , n - 1 , 1);此时ans里边记录的就是当前序列的逆序数之和

 

1、题目大意:

给定一串数字,求这一组数字的逆序数,而且这组数据可以改变,一次将前边的第一个数移到最后一个数的位置,构成新的数列,在诸多序列中,求出一个最小的逆序数

2、思路:

简单的线段树处理,单点更新即可,网上有一个很巧妙的处理最小逆序数的方法,

当把x放入数组的后面,此时的逆序数应该为x没放入最后面之前的逆序总数加上(n-x)再减去(x-1);sum = sum+(n-x[i])-(x[i]-1)。

此方法现在才明白,例如3 1 4 2 5,如果将3移到最后,那么比3大的数4,5的逆序数分别加1,即上式中的加上(n-x),而比3小的1,2的逆序数相比较以前分减1,即上式中的再减(x-1)

3、题目:

Minimum Inversion Number
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5831    Accepted Submission(s): 3547Problem 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 2Sample Output
16Author
CHEN, GaoliSource
ZOJ Monthly, January 2003 Recommend
Ignatius.L


 

4、代码:

#include<stdio.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
int x[5005];
int sum[5005*4];
void build(int l,int r,int rt)
{sum[rt]=0;if(l==r) return ;int m=(l+r)>>1;build(lson);build(rson);
}
int query(int L,int R,int l,int r,int rt)
{if(L<=l&&R>=r)return sum[rt];int m=(l+r)>>1,ret=0;if(L<=m)ret+=query(L,R,lson);if(R>=m+1)ret+=query(L,R,rson);return ret;
}
void update(int x,int l,int r,int rt)
{if(l==r){sum[rt]++;return ;}int m=(l+r)>>1;if(x<=m)update(x,lson);if(x>=m+1)update(x,rson);sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}
int main()
{int n,ans;while(scanf("%d",&n)!=EOF){ans=0;build(0,n-1,1);for(int i=1; i<=n; i++){scanf("%d",&x[i]);ans+=query(x[i],n-1,0,n-1,1);update(x[i],0,n-1,1);}int minsum=ans;for(int i=1; i<=n; i++){ans=ans+(n-x[i])-(x[i]+1);minsum=min(ans,minsum);}printf("%d\n",minsum);}return 0;
}
/*
10
1 3 6 9 0 8 5 7 4 2
*/


 

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