80 lines
2.1 KiB
C++
80 lines
2.1 KiB
C++
/*
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逆序对数量：
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归并排序解法 https://blog.csdn.net/m0_50299150/article/details/122729602
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树状数组解法 https://blog.csdn.net/m0_50299150/article/details/122729602
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权值线段树解法
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*/
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// 446 ms
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//此方法，vector+离散化+STL二分查找，在AcWing 可以正常通过，在洛谷挂一半，TLE,怀疑是lower_bound太慢
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#include <bits/stdc++.h>
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using namespace std;
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typedef long long LL;
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const int N = 5e5 + 10;
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int a[N];
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vector<int> nums;
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int n;
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struct node {
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int l, r, sum;
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} tr[N << 2];
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void build(int root, int l, int r) {
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tr[root] = {l, r, 0};
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if (l == r) return;
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int mid = (l + r) >> 1;
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build(root << 1, l, mid);
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build(root << 1 | 1, mid + 1, r);
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}
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void pushup(int root) {
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tr[root].sum = tr[root << 1].sum + tr[root << 1 | 1].sum;
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}
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void modify(int root, int pos, int val) {
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if (tr[root].l == pos && tr[root].r == pos) {
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tr[root].sum += val;
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return;
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}
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int mid = tr[root].l + tr[root].r >> 1;
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if (mid >= pos)
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modify(root << 1, pos, val);
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else
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modify(root << 1 | 1, pos, val);
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pushup(root);
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}
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int query(int root, int l, int r) {
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if (tr[root].l == l && tr[root].r == r) return tr[root].sum;
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int mid = tr[root].l + tr[root].r >> 1;
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if (mid < l)
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return query(root << 1 | 1, l, r);
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else {
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if (r <= mid)
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return query(root << 1, l, r);
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else
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return query(root << 1, l, mid) + query(root << 1 | 1, mid + 1, r);
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}
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}
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int find(int x) {
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return lower_bound(nums.begin(), nums.end(), x) - nums.begin() + 1;
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}
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int main() {
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//优化读入
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ios::sync_with_stdio(false);
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cin >> n;
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for (int i = 1; i <= n; i++) {
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cin >> a[i];
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nums.push_back(a[i]);
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}
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//离散化
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sort(nums.begin(), nums.end());
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nums.erase(unique(nums.begin(), nums.end()), nums.end());
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LL ans = 0;
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build(1, 1, nums.size() + 1);
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for (int i = 1; i <= n; i++) {
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int p = find(a[i]);
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ans += query(1, p + 1, nums.size() + 1);
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modify(1, p, 1);
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}
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cout << ans << endl;
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return 0;
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} |