整理手套

Description

Solution

可见手套的编号对于解决问题是没有关系的，于是我们对手套重新按出现时间编号。可见答案就是新序列的逆序对数量。
因为存在重复元素，所以采用归并排序来求逆序对。

#include<bits/stdc++.h>using namespace std;#define N 400001#define rep(i, a, b) for (int i = a; i <= b; i++)#define ll long longinline int read() {    int x = 0, flag = 1; char ch = getchar(); while (!isdigit(ch)) { if (!(ch ^ '-')) flag = -1; ch = getchar(); }    while (isdigit(ch)) x = (x << 1) + (x << 3) + ch - '0', ch = getchar(); return x * flag;}int n, a[N], b[N], t[N], tot;ll ans;void merge(int l, int r) {    if(!(l ^ r)) return;    int m = l + r >> 1; merge(l, m), merge(m + 1, r);    int p = l, q = m + 1, k = l;    while(p <= m && q <= r)        if(a[p] <= a[q]) t[k++] = a[p++];        else t[k++] = a[q++], ans += q - k;    while(p <= m) t[k++] = a[p++]; while(q <= r) t[k++] = a[q++];    rep(i, l, r) a[i] = t[i];}int main() {    n = read() << 1; rep(i, 1, n) if(!b[(a[i] = read())]) b[a[i]] = ++tot; rep(i, 1, n) a[i] = b[a[i]];    merge(1, n); cout << ans; return 0;}