HDU 1394 Minimum Inversion Number 线段树和逆序数的应用

地址：http://acm.hdu.edu.cn/showproblem.php?pid=1394

题意：给出n个数，通过移动，求最小的逆序数对

题解：

      1.主要代码：sum = sum + (n - 1 - arr2[i]) - arr2[i];      2.用线段树求初始数组的逆序数对，然后用上面的代码求出最小的逆序数对

#include<iostream>
#include<algorithm>
using namespace std;
#define lx (root<<1)
#define rx ((root<<1)|1)
#define mid ((left+right)>>1)
#define maxs 5010
int arr[maxs];
int arr2[maxs];
int temp[maxs];
int n,sum;
void put_up(int left ,int right){
int cur = 1;
int i = left;
int j = mid + 1;
while (i <= mid && j <= right){
if (arr[i] < arr[j])
temp[cur++] = arr[i++];
else{
temp[cur++] = arr[j++];
sum += mid - i + 1;
}
}
while (i<=mid)
temp[cur++] = arr[i++];
while (j<=right)
temp[cur++] = arr[j++];
for (int k = 1; k < cur;k++){
arr[left++] = temp[k];
}
}
int update(int root, int left, int right){
if (left == right)
return 0;
update(lx, left, mid);
update(rx, mid + 1, right);
put_up(left, right);
}
int main(){
while (scanf("%d", &n) != EOF){
sum = 0;
for (int i = 1; i <= n; i++){
scanf("%d", &arr[i]);
arr2[i] = arr[i];
}
update(1,1,n);
int mins = sum;
for (int i = 1; i < n; i++){
sum = sum + (n - 1 - arr2[i]) - arr2[i];
if (mins > sum)
mins = sum;
}
printf("%d\n", mins);
}
}