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
数据较小,可以暴力:
#include <iostream>
using namespace std;
const int MAX = 5e3 + 5;
int num[MAX];
int main()
{
ios::sync_with_stdio(false);
int N;
while (cin >> N){
for (int i = 1; i <= N; i++){
cin >> num[i];
}
int sum = 0;
for (int i = 1; i <= N; i++){
for (int j = 1; j < i; j++){
if(num[j] > num[i]){
sum ++;
}
}
}
int min = sum;
for (int i = 1; i < N; i++){
sum = sum - num[i] + (N - num[i] - 1);
if (min > sum){
min = sum;
}
}
cout << min << endl;
}
return 0;
}
线段树:
#include <iostream>
#include <algorithm>
using namespace std;
const int MAX = 2e5 +5;
int num[MAX], st[MAX << 2];
void PushUp(int rt){
st[rt] = st[rt << 1] + st[rt << 1 | 1];
}
void Build (int l, int r, int rt){
st[rt] = 0;
if (l == r){
return ;
}
int m = (l + r) >> 1;
Build(l, m, rt << 1);
Build(m + 1, r, rt << 1 | 1);
}
void Update(int p, int l, int r, int rt){
if (l == r){
st[rt]++;
return ;
}
int m = (l + r) >> 1;
if (p <= m){
Update(p, l, m, rt << 1);
}else {
Update(p, m + 1, r, rt << 1 | 1);
}
PushUp(rt);
}
int Query (int L, int R, int l, int r, int rt){
if (L <= l && r <= R){
return st[rt];
}
int m = (l + r) >> 1;
int res = 0;
if (L <= m){
res += Query(L, R, l, m, rt << 1);
}
if (m < R){
res += Query(L, R, m + 1, r, rt << 1 | 1);
}
return res; //这里不是pushup()
}
int main()
{
ios::sync_with_stdio(false);
int N;
while (cin >> N){
Build(0, N - 1, 1);
int res = 0;
for (int i = 0; i < N; i++){
cin >> num[i];
res += Query(num[i], N - 1, 0, N - 1, 1); //11
Update(num[i], 0, N - 1, 1); //22 这两行的顺序
}
int Min = res;
for (int i = 0; i < N; i++){
res = res - num[i] + (N - num [i] - 1);
Min = min(Min, res);
}
cout << Min << endl;
}
return 0;
}