HDU 1394 Minimum Inversion Number(线段树,单节点修改,区间求和)

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 21314    Accepted Submission(s): 12780

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

 

求 逆序数最小,  每次对头排到队尾;

类似树状数组,  线段树 建立 ,把 数的叶子作为每个数对应的位置, 然后根节点表示数目,  最后枚举所有的点 求逆序数,     (单节点增加 ,  区间求和)

逆序数 表达为  n-x-x-1;

代码:

#include <iostream>#include <stdio.h>#include <algorithm>#include <cmath>#include <cstring>#include <string>#include <queue>#include <stack>#include <stdlib.h>#include <list>#include <map>#include <set>#include <bitset>#include <vector>#define mem(a,b) memset(a,b,sizeof(a))#define findx(x) lower_bound(b+1,b+1+bn,x)-b#define FIN      freopen("input.txt","r",stdin)#define FOUT     freopen("output.txt","w",stdout)#define S1(n)    scanf("%d",&n)#define S2(n,m)  scanf("%d%d",&n,&m)#define Pr(n)     printf("%d\n",n)using namespace std;typedef long long ll;const double PI=acos(-1);const int INF=0x3f3f3f3f;const double esp=1e-6;const int maxn=1e6+5;const int MOD=1e9+7;const int mod=1e9+7;int dir[5][2]={0,1,0,-1,1,0,-1,0};ll inv[maxn*2],fac[maxn];ll gcd(ll a,ll b){ return b?gcd(b,a%b):a;}ll exgcd(ll a,ll b,ll &x,ll &y){if(!b){x=1;y=0;return a;}ll ans=exgcd(b,a%b,x,y);ll temp=x;x=y;y=temp-a/b*y;return ans;}ll lcm(ll a,ll b){ return b/gcd(a,b)*a;}ll qpow(ll x,ll n){ll res=1;for(;n;n>>=1){if(n&1)res=(res*x)%MOD;x=(x*x)%MOD;}return res;}void INV(){inv[1] = 1;for(int i = 2; i < maxn; i++) inv[i] = (MOD - MOD / i) * inv[MOD % i] % MOD;}void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){ x=1; y=0; d=a; }else{ ex_gcd(b,a%b,d,y,x); y-=x*(a/b);}}void Fac(){fac[0]=1;for(int i=1;i<=maxn;i++)fac[i]=(fac[i-1]*i)%MOD;}ll inv_exgcd(ll a,ll n){ll d,x,y;ex_gcd(a,n,d,x,y);return d==1?(x+n)%n:-1;}ll inv1(ll b){return b==1?1:(MOD-MOD/b)*inv1(MOD%b)%MOD;}ll inv2(ll b){return qpow(b,MOD-2);}ll cal(ll m,ll n){if(m<n)return 0;return (fac[m]*inv[fac[n]]%MOD)%MOD*inv[fac[m-n]]%MOD;}ll cals(ll m,ll n){if(m<n)return 0;return (fac[m]*inv1(fac[n])%MOD)%MOD*inv1(fac[m-n])%MOD;}int n;struct SegTree{int val;int addmark;}Seg[maxn];void PushUP(int root){Seg[root].val=Seg[root*2+1].val+Seg[root*2+2].val;}void build(int root,int begin,int end){    Seg[root].addmark=0;    Seg[root].val=0;if(begin==end)        return ;        //scanf("%d",&Seg[root].val);else{int mid=(begin+end)>>1;build(root*2+1,begin,mid);build(root*2+2,mid+1,end);PushUP(root);}}int query(int root,int begin,int end,int left,int right){if(left<=begin&&right>=end)return Seg[root].val;int mid=(begin+end)>>1;int ans=0;if(left<=mid) ans+= query(root*2+1,begin,mid,left,right);if(right>mid) ans+= query(root*2+2,mid+1,end,left,right);    //cout<<"ans="<<ans<<endl;return ans;}void insert(int root,int l,int r,int idex,int add){if(l==r){        Seg[root].val+=add;return;}int mid=(l+r)>>1;if(idex<=mid)insert(root*2+1,l,mid,idex,add);elseinsert(root*2+2,mid+1,r,idex,add);PushUP(root);}int x[maxn];int main(){    while(~S1(n))    {        mem(x,0);        build(1,0,n-1);        int sum=0;        for(int i=0;i<n;i++)        {            S1(x[i]);            sum+=query(1,0,n-1,x[i],n-1);            insert(1,0,n-1,x[i],1);        }        int ans=sum;        //cout<<"sum==="<<ans<<endl;        for(int i=0;i<n;i++)        {            sum+=n-x[i]-x[i]-1;            ans=min(ans,sum);        }        Pr(ans);    }    return 0;}

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