Minimum Inversion Number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 21957 Accepted Submission(s): 13098
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
思路:线段树区间最值,单点增减,求最小逆序数,先求出一个,然后可以发现公式,进而推出其他的解
实现代码:
#include#include #include using namespace std;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1const int M = 5555;int sum[M<<2];int n;void pushup(int rt){ 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 m = (l+r) >> 1; build(lson); build(rson);}void update(int p,int l,int r,int rt){ if(l == r){ sum[rt] ++; return ; } int m = (l + r) >> 1; if(p <= m) update(p,lson); else update(p,rson); pushup(rt);}int query(int L,int R,int l,int r,int rt){ if(L <= l && r <= R){ return sum[rt]; } int m = (l + r) >> 1; int ret = 0; if(L <= m) ret += query(L,R,lson); if(R>m) ret += query(L,R,rson); return ret;}int a[M];int main(){ int sum1; while(scanf("%d",&n)!=EOF){ build(0,n-1,1); int sum1 = 0; for(int i = 0;i < n;i++){ scanf("%d",&a[i]); sum1 += query(a[i]+1,n-1,0,n-1,1); update(a[i]+1,0,n-1,1); } int ans = sum1; for(int i = 0;i < n; i++){ sum1 = sum1 - a[i] + n - a[i] - 1; ans = min(ans,sum1); } printf("%d\n",ans); } return 0;}