BZOJ3925

发表于 | 阅读数

题意：可修改逆序对

开始没理清数据，越打越乱

理清了一波三维偏序，舒服了，自己代码没打错，只是思路错了。

我们先把该删的数删掉那么开始反向添加

比如

所以就三维偏序统计一个数的贡献就好了。

然后反向加就好了

代码 
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#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
const int N = 5e5 + 10;
typedef long long ll;
#define lowbit(i) i &(-i)
int tr[N << 2], n;
void update(int x, int w)
{
    while (x <= n)
    {
        tr[x] += w;
        x += lowbit(x);
    }
    return;
}
int quary(int x)
{
    int sum = 0;
    while (x)
    {
        sum += tr[x];
        x -= lowbit(x);
    }
    return sum;
}
struct node
{
    int pos, v, ti, id;
} p[N];
int m;
ll ans1[N], ans2[N], f[N];
bool cmpp(node a, node b)
{
    return a.ti < b.ti;
}
void cdq(int l, int r)
{
    if (l == r)
        return;
    int mid = (l + r) >> 1;
    cdq(l, mid);
    cdq(mid + 1, r);
    sort(p + l, p + mid + 1, cmpp);
    sort(p + 1 + mid, p + r + 1, cmpp);
    int pl = l, pr = mid + 1;
    while (pl <= mid)
    {
        //cout<<pl<<" "<<pr<<endl;
        while (pr <= r && p[pl].ti >= p[pr].ti)
            update(p[pr].pos, 1), pr++;
        ans1[p[pl].id] += quary(p[pl].pos - 1);
        pl++;
    }
    //分别统计
    for (int i = mid + 1; i < pr; i++)
        update(p[i].pos, -1);
    pl = l, pr = mid + 1;
    while (pr <= r)
    {
        //cout<<pl<<" "<<pr<<endl;
        while (pl <= mid && p[pl].ti <= p[pr].ti)
            update(p[pl].pos, 1), pl++;
        ans2[p[pr].id] += quary(n) - quary(p[pr].pos);
        pr++;
    }
    for (int i = l; i < pl; i++)
        update(p[i].pos, -1);
}
int main()
{
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++)
    {
        int x;
        scanf("%d", &x);
        p[x].pos = i;
        p[x].v = x;
    }
    for (int i = 1; i <= m; i++)
    {
        int x;
        scanf("%d", &x);
        p[x].ti = m - i + 1;
        p[x].id = i;
    }

    cdq(1, n);
    f[m + 1] = ans1[0] + ans2[0];
    f[m + 1] /= 2;
    //    for(int i=0;i<=m;i++)
    //    	cout<<ans1[i]<<' '<<ans2[i]<<endl;
    for (int i = m; i >= 1; i--)
        f[i] = f[i + 1] + ans1[i] + ans2[i];
    for (int i = 1; i <= m; i++)
        printf("%lld\n", f[i]);
}