HDOJ1394 Minimum Inversion Number(线段树)

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

101 3 6 9 0 8 5 7 4 2

 

Sample Output

16

 

问最小逆序数是多少~

AC代码：

#include "iostream"#include "cstdio"#include "cstring"#include "algorithm"using namespace std;#define lson l, m, rt << 1#define rson m + 1, r, rt << 1 | 1const int maxn = 5005;int sum[maxn << 2], x[maxn];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 updata(int p, int l, int r, int rt){if(l == r) {sum[rt]++;return;}int m = (l + r) >> 1;if(p <= m) updata(p, lson);else updata(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, ret = 0;if(L <= m) ret += query(L, R, lson);if(R > m) ret += query(L, R, rson);return ret;}int main(int argc, char const *argv[]){int n;while(scanf("%d", &n) != EOF) {build(0, n - 1, 1);int sum = 0;for(int i = 0; i < n; ++i) {scanf("%d", &x[i]);sum += query(x[i], n - 1, 0, n - 1, 1);updata(x[i], 0, n - 1, 1);}int ans = sum;for(int i = 0; i < n; ++i) {sum += n - x[i] - x[i] - 1;ans = min(ans, sum);}printf("%d\n", ans);}return 0;}