Hdu1394Minimum Inversion Number线段树

　　这个网上一搜一大堆，就是先求一个，其余的for一遍搞出来。

#include<stdio.h>
#include<stdlib.h>
#define max 5555
int sum[max * 4];
int min(int a, int b)
{
    if (a>b) return b;
    else return a;
}
void fuqin(int a)
{
    sum[a] = sum[a * 2] + sum[a * 2 + 1];
}
void build(int l, int r, int a)
{
    void fuqin(int);
    sum[a] == 0;
    if (l == r)
        return;
    int mid = (l + r) / 2;
    build(l, mid, a * 2);
    build(mid + 1, r, a * 2 + 1);
    fuqin(a);
}
void up(int q, int l, int r, int a)
{
    void fuqin(int);
    if (l == r)
    {
        sum[a]++;
        return;
    }
    int mid = (l + r) / 2;
    if (q <= mid) up(q, l, mid, a * 2);
    else up(q, mid + 1, r, a * 2 + 1);
    fuqin(a);

}
int ask(int L, int R, int l, int r, int a)
{
    if (L <= l&&r <= R)
        return sum[a];
    int mid = (l + r) / 2;
    int ans = 0;
    if (mid >= L) ans += ask(L, R, l, mid, a * 2);
    if (mid<R) ans += ask(L, R, mid + 1, r, a * 2 + 1);
    return ans;
}
void main()
{
    int n, i, x[5555], ans = 0, q;
    int min(int, int);
    void build(int, int, int);
    void tianjia(int, int, int, int);
    int ask(int, int, int, int, int);
    while (scanf("%d", &n) != EOF)
    {
        ans = 0;
        memset(sum, 0, sizeof(sum));
        build(0, n - 1, 1);
        for (i = 0; i<n; i++)
        {
            scanf("%d", &x[i]);
            ans += ask(x[i], n - 1, 0, n - 1, 1);
            up(x[i], 0, n - 1, 1);
        }
        q = ans;
        for (i = 0; i<n; i++)
        {
            q = q + (n - 1 - x[i]) - x[i];
            ans = min(q, ans);
        }
        printf("%d\n", ans);
    }
}