Minimum Inversion Number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
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.
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:给出一个含有n个数的序列,然后每次都把第一个数放到序列的最后面形成新的序列,求这n个序列中最小的逆序数是多少?
分析:由于数列中每一个元素的值都不同样,所以我们仅仅需求出原始序列的逆序数,然后依据这个逆序数递推出其它序列的逆序数。
比如序列 2 0 3 1 4 5 的逆序数是3。把2放到最后边以后。比2小的数(2个)的每一个数的逆序数减1,比2大的数(n-a[i]-1个)的逆序数不变。而2的逆序数变为比2大的数的个数(n-a[i]-1)。依据这个结论,我们就能够递推出其它序列的逆序数,进而求出最小值。
求原始序列的逆序数时用线段树来求,对于第i个数,它的逆序数等于已经插入到线段树中的比它大的数的个数。
#include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int N = 5005; #define lson l, mid, root<<1 #define rson mid+1, r, root<<1|1 int sum[N<<2], a[N]; void Push_Up(int root) { sum[root] = sum[root<<1] + sum[root<<1|1]; } void build_tree(int l, int r, int root) { sum[root] = 0; if(l == r) return; int mid = (l + r) >> 1; build_tree(lson); build_tree(rson); } void update(int p, int l, int r, int root) { if(l == r) { sum[root]++; return; } int mid = (l + r) >> 1; if(p <= mid) update(p, lson); else update(p, rson); Push_Up(root); } int Query(int L, int R, int l, int r, int root) { if(L <= l && r <= R) { return sum[root]; } 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, i; while(~scanf("%d",&n)) { build_tree(0, n-1, 1); int res = 0; for(i = 0; i < n; i++) { scanf("%d",&a[i]); res += Query(a[i], n-1, 0, n-1, 1); //查询已经插入到线段树中的数有多少个数比a[i]大 update(a[i], 0, n-1, 1); //把这个数插入到线段树中 } int ans = res; for(i = 0; i < n; i++) { res += (n - a[i] - 1) - a[i]; ans = min(ans, res); } printf("%d ", ans); } return 0; }