Minimum Inversion Number HDU 1394

题目链接:点我

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    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

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题意:

求循环最小逆序对数

思路:

线段树单点更新,区间查询.每次将一个数插入相应节点前,先区间查询这个比这个数大的数的个数,因为他们比这个数先出现而且比这个数大,所以构成逆序对.问题的难点在与如何求循环的逆序对数,假设我们把数字k移到到最后去,那么逆序对增加的个数为比k大的数的个数即:n - k - 1,相应的减少的数的个数为小于k的数的个数,即: k,最后递推即可.

代码:

#include<cstdio>#include<algorithm>#include<cstring>#include<iostream>#include<cmath>using namespace std;int b[5000+10];int a[20000];void pushup(int cur){    a[cur] = a[cur << 1] + a[cur << 1|1];//维护逆序对数的总和}void build(int cur, int l, int r){    a[cur] = 0;    if(l == r)        return ;    int mid = (l + r) >> 1;    build( cur << 1, l ,mid);    build( cur << 1|1,mid + 1, r);}int quiry(int ll, int rr, int l, int r, int cur){    if (ll <= l && rr >= r)//如果当前区间在要查询区间内,直接返回.        return a[cur];    int mid = (l + r) >> 1;    int ans=0;    if (ll <= mid) ans += quiry( ll, rr, l, mid, cur << 1);    if (rr > mid) ans += quiry( ll , rr, mid+1, r, cur << 1|1);    return ans;}void update(int key, int l, int r, int cur){    if(l == r){        a[cur]++;        return ;    }    int mid = (l + r) >> 1;    if(key <= mid) update( key, l, mid, cur << 1);    else update( key, mid + 1, r, cur << 1|1);    pushup(cur);}int main(){    int n;    while(scanf("%d", &n) != EOF){        build(1,0,n-1);        int sum = 0;        for(int i = 0;i < n; ++ i){            scanf("%d", &b[i]);            sum += quiry( b[i], n-1, 0, n-1, 1);            update( b[i], 0 , n-1, 1);        }        int ans = sum;        for(int i = 0; i < n; ++ i){//递推求最小逆序对数            sum += n - b[i] - b[i] - 1;            ans = min( ans, sum);        }        printf("%d\n", ans);    }    return 0;}