HDU 1394 线段树求逆序对

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25905    Accepted Submission(s): 15250

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

题意：一个长度为n 的 0-n-1 的全排列  你可以进行若干次操作 把第一个数放到最后面 形成一个新的序列(最多n个）问所有序列中逆序对最少的数量。

解析：逆序对很好求 对于每一个数看它前面有多少个比它大的记为sum[ a[i] ] sigmasum[] 就是逆序的数量 每个位置都如此，线段树就是每次查询[a[i],n-1] 的个数；

假如当前序列的逆序对是ans 那么把当前第一个数a[i] 放到最后面 后面对于每个数[0，a[i]) sum[ ] 就会少一次 同样 sum[ a[i] ] = n-1-a[i]

所以 操作之后就是 ans+=n-a[i]+n-1-a[i]。取最大值就好了

#include <bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define all(a) (a).begin(), (a).end()
#define fillchar(a, x) memset(a, x, sizeof(a))
#define huan printf("
")
#define debug(a,b) cout<<a<<" "<<b<<" "<<endl
#define ffread(a) fastIO::read(a)
using namespace std;
typedef long long ll;
const int maxn = 1e4+10;
const int inf = 0x3f3f3f3f;
const int mod = 100000007;
const double epx = 1e-6;
const double pi = acos(-1.0);
//head-----------------------------------------------------------------
int sum[maxn*4];
void PushUp(int rt)
{
    sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}
void build(int l,int r,int rt)
{
    if(l==r)
    {
        sum[rt]=0;
        return;
    }
    int mid=(l+r)>>1;
    build(l,mid,rt<<1);
    build(mid+1,r,rt<<1|1);
    PushUp(rt);
}
void update(int x,int val,int l,int r,int rt)
{
    if(l==x&&r==x)
    {
       sum[rt]+=val;
       return;
    }
    int mid=(l+r)>>1;
    if(x<=mid)
        update(x,val,l,mid,rt<<1);
    else
        update(x,val,mid+1,r,rt<<1|1);
    PushUp(rt);
}
int query(int L,int R,int l,int r,int rt)
{
    if(L<=l&&R>=r)
    {
        return sum[rt];
    }
    int mid=(l+r)>>1;
    int ans=0;
    if(L<=mid)
        ans+=query(L,R,l,mid,rt<<1);
    if(R>mid)
        ans+=query(L,R,mid+1,r,rt<<1|1);
    return ans;
}
int a[maxn];
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        fillchar(sum,0);
        int ans=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            update(a[i]+1,1,1,n,1);  //出于方便从1开始
            ans+=query(a[i]+2,n,1,n,1);
        }
        int ret=inf;
        for(int i=1;i<=n;i++)
        {
            ans+=n-2*a[i]-1;
            ret=min(ans,ret);
        }
        printf("%d
",ret);
    }
}

复习一下线段树基础。。