POJ 1804 Brainman

题目链接：POJ 1804 Brainman

题目大意：

题解：
在归并排序的过程中记录逆序对个数。

#include <algorithm>
#include <iostream>
using namespace std;

int ans, s[1005], t[1005], n, T;

void merge(int l, int r) {
    int p, q, mid, len;
    if (l == r) return;
    mid = l + r >> 1;
    merge(l, mid);
    merge(mid + 1, r);
    p = l;
    q = mid + 1;
    len = r - l + 1;
    for (int i = 0; i < len; ++i) {
        if ((q > r) || (s[p] <= s[q] && p <= mid)) {
            t[l + i] = s[p++];
        } else {
            t[l + i] = s[q++];
            ans += mid - p + 1;
        }
    }
    for (int i = l; i <= r; ++i) {
        s[i] = t[i];
    }
}

int main() {
    cin >> T;
    for (int t = 1; t <= T; ++t) {
        cin >> n;
        for (int i = 1; i <= n; ++i) {
            cin >> s[i];
        }
        ans = 0;
        merge(1, n);
        cout << "Scenario #" << t << ":" << endl << ans << endl << endl;
    }
    return 0;
}