POJ2299 Ultra-QuickSort

Posted 2021-01-13 captain1

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树状数组求逆序对 + 离散化？

我好像永远都记不住怎么用树状数组求逆序对……以前我记得都是正着写的，这次怎么得倒着写才能过……

然后本题a的范围极大，但是n的范围不大，500000，需要先手离散化一波，之后就可以正常按照树状数组操作了。然后在query的时候注意要-1.

我们来看一下代码。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<queue>
#include<cstring>
#include<utility>
#include<map>
#define pr pair<int,int>
#define mp make_pair
#define fi first
#define sc second
#define rep(i,a,n) for(int i = a;i <= n;i++)
#define per(i,n,a) for(int i = n;i >= a;i--)
#define enter putchar(‘
‘)
#define lowbit(x) x & (-x)
using namespace std;
typedef long long ll;
const int M = 500005;
const int N = 32005;

int read()
{
    int ans = 0,op = 1;
    char ch = getchar();
    while(ch < ‘0‘ || ch > ‘9‘)
    {
    if(ch == ‘-‘) op = -1;
    ch = getchar();
    }
    while(ch >=‘0‘ && ch <= ‘9‘)
    {
    ans *= 10;
    ans += ch - ‘0‘;
    ch = getchar();
    }
    return ans * op;
}

int n,a[M],c[M],tot,d[M];
ll ans;

void add(int x)
{
    while(x <= M-5) a[x]++,x += lowbit(x);
}

int query(int x)
{
    int cur = 0;
    while(x) cur += a[x],x -= lowbit(x);
    return cur;
}

void clear()
{
    tot = ans = 0;
    memset(a,0,sizeof(a));
}

int main()
{
    while(1)
    {
    n = read();
    if(n == 0) break;
    clear();
    rep(i,1,n) c[i] = read(),d[i] = c[i];
    sort(c+1,c+1+n);
    tot = unique(c+1,c+1+n) - c - 1;
    per(i,n,1)
    {
        int k = lower_bound(c+1,c+1+tot,d[i]) - c;
        //printf("#%d
",k);
        ans += query(k-1),add(k);
    }
    printf("%lld
",ans);
    }
    return 0;
}