HDU 1394 Minimum Inversion Number （线段树：单点增减求和）

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18375    Accepted Submission(s): 11156

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

Source

ZOJ Monthly, January 2003

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题解：问你不断地将开头的数放到结尾。求其中最小的逆序数。

线段树求出最初的逆序数。复杂度O（nlogn），然后O(n)查询其他解。

AC代码：

#pragma comment(linker, "/STACK:102400000,102400000")//#include<bits/stdc++.h>#include <stdio.h>#include <string.h>#include <algorithm>#include <iostream>#include <cstring>#include <vector>#include <map>#include <cmath>#include <queue>#include <set>#include <bitset>#include <iomanip>#include <list>#include <stack>#include <utility> using namespace std;typedef long long ll;typedef unsigned long long ull;typedef pair<int,int> pii;typedef vector<int> vi;const double eps = 1e-8;  const int INF = 1e9+7; const ll inf =(1LL<<62) ;const int MOD = 1e9 + 7;  const ll mod = (1LL<<32);const int N =1e6+6; const int M=100010;const int maxn=55555;#define mst(a) memset(a, 0, sizeof(a))#define M_P(x,y) make_pair(x,y)#define in freopen("in.txt","r",stdin) #define rep(i,j,k) for (int i = j; i <= k; i++)  #define per(i,j,k) for (int i = j; i >= k; i--)  #define lson  l , mid , rt << 1    #define rson  mid + 1 , r , rt << 1 | 1  const int lowbit(int x) { return x&-x; }//const int lowbit(int x) { return ((x)&((x)^((x)-1))); } int read(){ int v = 0, f = 1;char c =getchar();while( c < 48 || 57 < c ){if(c=='-') f = -1;c = getchar();}while(48 <= c && c <= 57) v = v*10+c-48, c = getchar();return v*f;}int sum[maxn<<2];int x[maxn];void pushup(int rt) //把当前结点的信息更新到父结点{//线段树是用数组来模拟树形结构//对于每一个节点rt,左子节点为 2*rt (一般写作rt<<1)右子节点为 2*rt+1（一般写作rt<<1|1） sum[rt] = sum[rt<<1] + sum[rt<<1|1]; }void build(int l,int r,int rt){sum[rt] = 0;if(l==r) return ; int mid=(l+r)>>1;build(lson);//递归构造左子树build(rson);//递归构造右子树pushup(rt); //更新求和 }void update(int p, int l, int r, int rt)//单点增 {if(l==r){sum[rt] ++; return ;}int mid=(l+r)>>1;if(p<=mid) update(p,lson);else update(p,rson);pushup(rt);}int query(int L,int R,int l,int r,int rt) //区间求逆序数{if(L <= l && r <= R){return sum[rt];}int mid = (l+r)>>1;int ans = 0;if(L <= mid) ans+=query(L,R,lson);if(R > mid) ans+=query(L,R,rson);return ans; }int main(){int n;while(~scanf("%d",&n)){build(0,n-1,1);int sum = 0;for(int i=0; i<n; i++){scanf("%d",&x[i]);sum+=query(x[i],n-1,0,n-1,1);update(x[i],0,n-1,1);}int ans = sum;for(int i=0;i<n;i++){sum=sum-x[i]; //减少了的逆序数 sum+=n-x[i]-1; //增加了的逆序数 ans=min(ans,sum);}cout<<ans<<endl; }     return 0;}