【LG1393】动态逆序对

题面

洛谷

题解

\(CDQ\)分治，按照时间来分治

应为一个删除不能对前面的操作贡献，所以考虑一个删除操作对它后面时间的操作的贡献

用上一个答案减去次贡献即可

代码

#include <iostream>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cmath>

#include <algorithm>

#include <vector>

using namespace std;

namespace IO {

    const int BUFSIZE = 1 << 20;

    char ibuf[BUFSIZE], *is = ibuf, *it = ibuf;

    inline char gc() {

        if (is == it) it = (is = ibuf) + fread(ibuf, 1, BUFSIZE, stdin);

		return *is++;

    }

}

inline int gi() {

    register int data = 0, w = 1;

    register char ch = 0;

    while (ch != '-' && (ch > '9' || ch < '0')) ch = IO::gc();

    if (ch == '-') w = -1 , ch = IO::gc();

    while (ch >= '0' && ch <= '9') data = data * 10 + (ch ^ 48), ch = IO::gc();

    return w * data;

}

const int MAX_N = 40005;

struct Node { int x, y, z, s; } t[MAX_N];

bool cmp_x(Node a, Node b) { return a.x < b.x; }

bool cmp_y(Node a, Node b) { return a.y < b.y; }

int N, M, ans, X[MAX_N], a[MAX_N], b[MAX_N], c[MAX_N], d[MAX_N];

inline int lb(int x) { return x & -x; }

void add(int x, int v) { while (x <= N) c[x] += v, x += lb(x); }

int sum(int x) { int res = 0; while (x > 0) res += c[x], x -= lb(x); return res; }

void Div(int l, int r) {

	if (l == r) return ;

	int mid = (l + r) >> 1;

	Div(l, mid); Div(mid + 1, r);

	int j = mid + 1;

	for (int i = l; i <= mid; i++) {

	    for ( ; j <= r && t[j].y < t[i].y; ) add(t[j].z, 1), ++j;

	    t[i].s += sum(N) - sum(t[i].z);

    }

    for (int i = mid + 1; i < j; i++) add(t[i].z, -1);

    j = r;

    for (int i = mid; i >= l; i--) {

        for ( ; j > mid && t[j].y > t[i].y; ) add(t[j].z, 1), --j;

        t[i].s += sum(t[i].z - 1);

    }

    for (int i = r; i > j; i--) add(t[i].z, -1);

    inplace_merge(&t[l], &t[mid + 1], &t[r + 1], cmp_y);

}

int main () {

    N = gi(), M = gi();

    for (int i = 1; i <= N; i++) X[i] = a[i] = gi();

    sort(&X[1], &X[N + 1]);

    for (int i = 1; i <= N; i++) a[i] = lower_bound(&X[1], &X[N + 1], a[i]) - X;

    for (int i = N; i >= 1; i--) ans += sum(a[i] - 1), add(a[i], 1);

    for (int i = 1; i <= N; i++) t[i].x = N + 1, t[i].y = i, t[i].z = a[i];

    for (int i = 1; i <= M; i++) {

        d[i] = gi();

        t[d[i]].x = i;

    }

    sort(&t[1], &t[N + 1], cmp_x);

	memset(c, 0, sizeof(c));

	Div(1, N);

	for (int i = 1; i <= N; i++) c[t[i].y] = t[i].s;

	printf("%d ", ans);

	for (int i = 1; i <= M; i++) printf("%d ", ans -= c[d[i]]);

    return 0;

}