HDU1394用线段树求逆序数

题意：一个由0..n-1组成的序列，每次可以把队首的元素移到队尾，

求形成的n个序列最小逆序对数目

算法：将元素依次插入线段树，每次增加的逆序对数为比它大的已经插入的

数的个数，可以用线段树维护，由于元素值为0..n,每次移动可求出增减

逆序对的数量更新。

#include <stdio.h>
#define MAXN 100000
#define ROOT 1

struct node
{
int left,right,sum;
}t[MAXN];

int val[MAXN];
int n;

void build(int p, int left, int right)
{
int m;
t[p].left = left;
t[p].right = right;
t[p].sum = 0;
if (left == right)
return;
m = (left + right) / 2;
build(p*2, left, m);
build(p*2+1, m+1, right);
}

void update(int p, int goal, int add)
{
t[p].sum += add;
if (t[p].left == t[p].right)
return;
int m = (t[p].left + t[p].right) / 2;
if (goal <= m)
update(p*2, goal, add);
if (goal > m)
update(p*2+1, goal, add);
}

int getsum(int p, int left, int right)
{
if (left > right)
return 0;
if (t[p].left == left && t[p].right == right)
return t[p].sum;
int m = (t[p].left + t[p].right) / 2;
if (right <= m)
return getsum(p*2, left, right);
else if (left > m)
return getsum(p*2+1, left, right);
else return getsum(p*2, left, m) + getsum(p*2+1, m + 1, right);
}

int main()
{
while (scanf("%d", &n) == 1)
{
build(ROOT, 0, n - 1);
int i,sum = 0,ans;
for (i = 1; i <= n; i++)
{
scanf("%d", &val[i]);
sum += getsum(ROOT, val[i], n - 1);
update(ROOT, val[i], 1);
}
ans = sum;
for (i = 1; i <= n; i++)
{
sum = sum + (n - val[i] - 1) - val[i];
ans = sum < ans ? sum : ans;
}
printf("%d\n", ans);
}
return 0;
}