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);

    }

}