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
题意：每次能够将第一个放到最后一个，求这个过程中最小的逆序数和
思路：线段树的点更新，首先读入序列的时候就能够算出一个逆序数，找找比它大的数的个数，
之后将第一个a[0]放到最后一个的时候，每次降低的是a[0],添加的是(n-a[0]-1)
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#define lson(x) ((x) << 1)
#define rson(x) ((x) << 1 | 1)
using namespace std;
const int MAXN = 5005;
const int ROOT = 1;
const int inf = 0x3f3f3f3f;
// single point updated
struct seg{
int w;
}; struct segment_tree {
seg node[MAXN * 4]; void update(int pos) {
node[pos].w = node[lson(pos)].w + node[rson(pos)].w;
} void build(int l, int r, int pos) {
if (l == r) { node[pos].w = 0; return; }
int m = l + r >> 1;
build(l, m, lson(pos));
build(m + 1, r, rson(pos));
update(pos);
} void modify(int l, int r, int pos, int x, int y) {
if (l == r) { node[pos].w = 1; return; }
int m = l + r >> 1;
if (x <= m) modify(l, m, lson(pos), x, y);
else modify(m + 1, r, rson(pos), x, y);
update(pos);
} int query(int l, int r, int pos, int x, int y) {
if (x <= l && r <= y) return node[pos].w;
int m = l + r >> 1;
int res = 0;
if (x <= m) res += query(l, m, lson(pos), x, y);
if (y > m) res += query(m + 1, r, rson(pos), x, y);
return res;
} int remove(int l, int r, int pos, int x) {
if (l == r) { node[pos].w = 0; return l; }
int m = l + r >> 1;
int res;
if (x < node[lson(pos)].w) res = remove(l, m, lson(pos), x);
else res= remove(m + 1, r, rson(pos), x - node[lson(pos)].w);
update(pos);
return res;
}
} tree; int main() {
int n;
int a[MAXN];
while (scanf("%d", &n) != EOF) {
tree.build(1, n, 1);
int cnt = 0;
for (int i = 0; i < n; i++) {
scanf("%d", &a[i]);
cnt += tree.query(1, n, 1, a[i]+1, n);
tree.modify(1, n, 1, a[i]+1, 1);
}
int ans = cnt;
for (int i = 0; i < n-1; i++) {
cnt += (n - 2*a[i] - 1);
ans = min(ans, cnt);
}
printf("%d\n", ans);
}
return 0;
}