ccnu-线段树-单点更新3-C

C - 单点更新3
Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u
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Status
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

代码如下：

【还是模板题，~T^T呜呜呜~没把题意读清楚，搞得理解了半天，讨厌死了~话说用线段树比暴力快了100倍呢⊙﹏⊙】

#include <cstdio>
#include <algorithm>
using namespace std;

#define lson l , m , rt << 1
#define rson m + 1 , r , rt << 1 | 1
const int maxn = 5010;
int sum[maxn<<2];
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 update(int p,int l,int r,int rt) {
	if (l == r) {
		sum[rt] ++;
		return ;
	}
	int m = (l + r) >> 1;
	if (p <= m) update(p , lson);
	else update(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;
	int ret = 0;
	if (L <= m) ret += query(L , R , lson);
	if (R > m) ret += query(L , R , rson);
	return ret;
}
int x[maxn];
int main() {
	int n;
	while (~scanf("%d",&n)) {
		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);
			update(x[i] , 0 , n - 1 , 1);
		}
		int ret = sum;
		for (int i = 0 ; i < n ; i ++) {
			sum += n - x[i] - x[i] - 1;
			ret = min(ret , sum);
		}
		printf("%d\n",ret);
	}
	return 0;
}