线段数组入门题题解

A - Ultra-QuickSort

本题要求逆序对，可以用归并排序或树状数组去求。
题解：离散化+树状数组。
将每个数值压入树状数组中，然后将离散化好的数组，倒叙进行树状数组的更新和查询，ans加上当前压入树状数组的比他小的数据的个数。
ACcode：

/*
 * @Author: NEFU_马家沟老三
 * @LastEditTime: 2020-06-22 15:48:33
 * @优快云 blog: https://blog.youkuaiyun.com/acm_durante
 * @E-mail: 1055323152@qq.com
 * @ProbTitle: 
 */
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
#include <stack>
#include <bitset>
#include <cstdlib>
#include <cmath>
#include <set>
#include <list>
#include <deque>
#include <map>
#include <queue>
#include <iomanip>
using namespace std;
typedef long long ll;
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define per(i, a, n) for (int i = n; i >= a; i--)
#define lowbit(x) ((x) & -(x))
const double PI = acos(-1.0);
int tree[500010], n, num[500010];
ll ans;
struct node
{
    int val, pos;
} a[500010];

void update(int x)
{
    while (x <= n)
    {
        tree[x]++;
        x += lowbit(x);
    }
}

ll query(int x)
{
    ll res = 0;
    while (x > 0)
    {
        res += (ll)tree[x];
        x -= lowbit(x);
    }
    return res;
}

bool cmp(node x, node y)
{
    if (x.val != y.val)
        return x.val < y.val;
}

int main()
{
    while (~scanf("%d", &n) && n)
    {
        memset(tree, 0, sizeof(tree));
        ans = 0;
        rep(i, 1, n)
        {
            scanf("%d", &a[i].val);
            a[i].pos = i;
        }
        sort(a + 1, a + 1 + n, cmp);
        for (int i = 1; i <= n; i++) //离散化
        {
            num[a[i].pos] = i;
        }
        per(i, 1, n)
        {
            update(num[i]);
            ans += query(num[i] - 1);
        }
        printf("%lld\n", ans);
    }
    return 0;
}

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