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
先百科一下逆序数的概念： 在一个排列中，如果一对数的前后位置与大小顺序相反，即前面的数大于后面的数，那么它们就称为一个 逆序 。一个排列中逆序的总数就称为这个排列的 逆序数 。逆序数为 偶数 的排列称为 偶排列 ；逆序数为奇数的排列称为奇排列 。如2431中，21，43，41，31是逆序，逆序数是4，为偶排列。
逆序数计算方法是：在逐个元素读取原始数列时，每次读取都要查询当前已读取元素中大于当前元素的个数，加到sum里，最后得到的sum即为原始数列的逆序数。
现在来说一下求解思路：
输入数组时，每输入一个数，就for(j=0;j<i-1;j++)比较大小，这样用sum把逆序数统计出来，其实这里才是暴力，至于后面的就很巧妙，公式很容易推出，因为题目总是把第一个数移到最后一个位置，所以原来比它小的数（和它构成逆序）在移动之后就不是逆序了，而原来比它大的数（不和它构成逆序）在移动之后就是逆序了，这样sum就变化了：Sum=sum-(low[a[i]])+(up[a[i]]); 显然在序列0,1,2,…..n-1中比a[i]小的数的个数是 Low[a[i]]=a[i]; 比a[i]大的数的个数是up[a[i]]=n-a[i]-1; 题目要求是循环移动n次，那么只要写个for，把a[0],a[1],a[2]……a[n-1]都移动一遍，sum进行n次上面的公式运算，同时记录最小值，就是最小逆序数了。
接下来是本馒头暴力AC的代码，数据为5000时的耗时为109ms，应该庆幸数据小= =
#include<cstdio>
#include<cstring>
#include<queue>
#include<cmath>
#include<algorithm>
using namespace std;
#define INF 0x3f3f3f3f
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
const int MX = 5050;
int num[MX];
int n;
int main() {
//freopen("input.txt", "r", stdin);
while (scanf("%d", &n) != EOF) {
int sum = 0;
for (int i = 0; i < n; i++) {
scanf("%d", &num[i]);
for (int j = 0; j < i; j++) {
if (num[j] > num[i]) sum++;
}
}
int Min = INF;
for (int i = 0; i < n; i++) {
sum = sum - num[i] + n - num[i] - 1;
Min = min(sum, Min);
}
printf("%d
", Min);
}
return 0;
}
由于最近在学习线段树，所以这道题也用线段树优化一下，主要是优化算出sum的那一段，使得整体算法时间复杂度减低到nlongn，对数默认以二为底，速度优化到62ms，以下是AC代码：
#include<cstdio>
#include<cstring>
#include<queue>
#include<cmath>
#include<algorithm>
using namespace std;
#define INF 0x3f3f3f3f
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
const int MX = 5050;
int num[MX];
int n;
int sum[MX<<2];
void push_up(int rt) {
sum[rt] = sum[rt<<1] + sum[rt<<1|1];
}
void build(int l, int r, int rt) {
if (l == r) {
sum[rt] = 0;
return ;
}
int m = (l + r)>>1;
build(lson);
build(rson);
push_up(rt);
}
void update(int pos, int l, int r, int rt) {
if (l == r) {
sum[rt]++;
return ;
}
int m = (l + r)>>1;
if (pos <= m) update(pos, lson);
else update(pos, rson);
push_up(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 main() {
//freopen("input.txt", "r", stdin);
while (scanf("%d", &n) != EOF) {
memset(sum, 0, sizeof(sum));
int s = 0;
build(1, n