题目传送门
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
题意
现在有一个数组，含有n个数（为0-（n-1）的一个排列），依次把第一个数，放到数组最后面，求其中逆序数数目最小是多少？
题解
-
先求初始状态的逆序数
我们先建立一颗空树，然后一个一个将节点插入树中，把数组中数的大小看成在树中的位置，然后每插入一个节点，就统计大于这个数的数目，即产生的逆序数的数目，再更新该位置的数值为1.这样就求出了初始状态下的逆序数sum了。 -
然后要求出最小的逆序数
因为数组是0到 n-1 的一个排列，所以如果当你把第一个数从第一个放到最后一个的时候，逆序数就会减少 a[i] 个，同时增加 n - (a[i]+1) 个。所以我们只需要枚举依次把第一个数放到最后位置，所产生的逆序数 sum - a[i] + n - a[i] - 1 取最小值即可。
AC-Code
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <vector>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <fstream>
#include <utility>
using namespace std;
typedef long long ll;
typedef pair<int, int> Pii;
#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);const int INF = 0x7fffffff;
const int MOD = 1e4;
const int MAXN = 5e3 + 7;struct node {
int left;int right;int data;
}tree[MAXN << 2];
int n;
void PushUp(int rt) {
tree[rt].data = tree[rt << 1].data + tree[rt << 1 | 1].data;
}void build(int rt, int l, int r) {
tree[rt].left = l;tree[rt].right = r;tree[rt].data = 0;if (l == r) {
return;}int mid = (l + r) >> 1;build(rt << 1, l, mid);build(rt << 1 | 1, mid + 1, r);PushUp(rt);
}
void updata(int rt, int id) {
if (tree[rt].left == tree[rt].right) {
tree[rt].data++;return;}int mid = (tree[rt].left + tree[rt].right) >> 1;if (id <= mid) updata(rt << 1, id);if (id > mid) updata(rt << 1 | 1, id);PushUp(rt);
}
int Query(int rt, int l, int r) {
if (l <= tree[rt].left && tree[rt].right <= r) {
return tree[rt].data;}int ans = 0;int mid = (tree[rt].left + tree[rt].right) >> 1;if (l <= mid) ans += Query(rt << 1, l, r);if (r > mid) ans += Query(rt << 1 | 1, l, r);return ans;
}
int a[MAXN];
int main() {
ios;while (cin >> n) {
build(1, 1, n);int sum = 0;for (int i = 1; i <= n; i++) {
cin >> a[i];sum += Query(1, a[i] + 1, n);updata(1, a[i] + 1);}int ans = sum;for (int i = 1; i <= n; i++) {
sum = sum - a[i] + n - a[i] - 1;ans = min(ans, sum);}cout << ans << endl;}return 0;
}