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TangDou/AcWing/P1908_2.cpp

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