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 1struct 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;}