HDU-1394 Minimum Inversion Number（线段树求逆序数）
HDU-1394 Minimum Inversion Number（线段树求逆序数）
Minimum Inversion Number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 15675 Accepted Submission(s): 9569
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
这道题目有两个地方可以留意一下。 首先求一个数列的逆序数 首先将这个数列排序，然后从最后一个数，边询问边更新 第二，根据题意，将第一个数放到最后一个，形成的新数列应该怎么求逆序数。其实可以递推，和没有变化前的数列比较逆序数增加了(n-1-a[i])-a[i]
#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
using namespace std;
#define MAX 5000
int segTree[MAX*4+5];
int n;
int a[MAX];
int b[MAX];
int c[MAX];
void PushUp(int node)
{
segTree[node]=segTree[node<<1]+segTree[node<<1|1];
}
void build(int node,int begin,int end)
{
if(begin==end)
{
segTree[node]=0;
return;
}
int m=(begin+end)>>1;
build(node<<1,begin,m);
build(node<<1|1,m+1,end);
PushUp(node);
}
void Update(int node,int begin,int end,int ind,int num)
{
if(begin==end)
{
segTree[node]+=num;
return;
}
int m=(begin+end)>>1;
if(ind<=m)
Update(node<<1,begin,m,ind,num);
else
Update(node<<1|1,m+1,end,ind,num);
PushUp(node);
}
int Query(int node,int begin,int end,int left,int right)
{
if(left<=begin&&end<=right)
return segTree[node];
int ret;
int m=(begin+end)>>1;
ret=0;
if(left<=m)
ret+=Query(node<<1,begin,m,left,right);
if(right>m)
ret+=Query(node<<1|1,m+1,end,left,right);
return ret;
}
int cmp(int x,int y)
{
return x<y;
}
int main()
{
int sum;
while(scanf("%d",&n)!=EOF)
{
sum=0;
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
c[i]=a[i];
b[a[i]]=i;
}
build(1,1,n);
sort(a+1,a+n+1,cmp);
for(int i=n;i>=1;i--)
{
sum+=Query(1,1,n,1,b[a[i]]);
Update(1,1,n,b[a[i]],1);
}
int num;
num=sum;
int ans;
ans=sum;
for(int i=1;i<n;i++)
{
num+=-c[i]+(n-1-c[i]);
if(ans>num)
ans=num;
}
printf("%d\n",ans);
}
return 0;腾讯云开发者
