【HDU1394】Minimum Inversion Number（线段树）

大意：n次操作原串查询逆序数，求出所有串中最小的逆序数。

求逆序数属于线段树的统计问题，建立空树，每次进行插点时进行一次query操作即可。n次操作可以套用结论：如果是0到n的排列，那么如果把第一个数放到最后，对于这个数列，逆序数是减少a[i],而增加n-1-a[i]。

#include <iostream>
#include <cstdio>
using namespace std;

const int maxn = 5e3 + 10;
int sumv[maxn << 2];

void PushUp (int rt) {
    sumv[rt] = sumv[rt * 2] + sumv[rt * 2 +1];
}

void build (int l, int r, int rt) {
    sumv[rt] = 0;
    if (l == r) {
        return ;
    }
    int m = (l + r) / 2;
    build (l, m, rt * 2);
    build (m + 1, r, rt * 2 + 1);
}

void update (int p, int l, int r, int rt) {
    if (l == r) {
        sumv[rt] ++;
        return ;
    }
    int m = (l + r) / 2;
    if (p <= m) {
        update (p, l, m, rt * 2);
    }
    if (p > m) {
        update (p, m + 1, r, rt * 2 + 1);
    }
    PushUp (rt);
}

int query (int L, int R, int l, int r, int rt) {
    if (L <= l && r <= R) {
        return sumv[rt];
    }
    int m = (l + r) / 2;
    int ret = 0;
    if (L <= m) {
        ret += query (L, R, l, m, rt * 2);
    }
    if (R > m) {
        ret += query (L, R, m + 1, r, rt * 2 + 1);
    }
    return ret;
}

int x[maxn];

int main () {
    int n;
    while (~scanf ("%d", &n)) {
        build (0, n - 1, 1);
        int sum = 0;
        for (int i = 0 ; i < n; ++ i) {
            scanf ("%d", &x[i]);
            sum += query (x[i], n - 1, 0, n - 1, 1);
            update (x[i], 0, n - 1, 1);
        }
        int res = sum;
        for (int i =  0; i < n; ++ i) {
            sum = sum + n - x[i] - 1 - x[i];
            res = min (res, sum);
        }
        printf ("%d
", res);
    }
    return 0;
}