重排文档(Reordering the Documents) 题解

思路分析

按题目要求使用对应数据结构与算法模拟/递推求解，注意数据范围边界与时间复杂度。

参考代码

#include <iostream>
#include <vector>
#include <numeric>
using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n, m;
    if (!(cin >> n >> m)) return 0;
    vector<int> a(n + 1);
    for (int i = 1; i <= n; ++i) cin >> a[i];
    vector<int> p(n + 1), col(n + 1, 0);
    iota(p.begin(), p.end(), 0);
    auto find = [&](auto& self, int x) -> int {
        if (p[x] == x) return x;
        int r = self(self, p[x]);
        col[x] ^= col[p[x]];
        return p[x] = r;
    };
    for (int i = 1; i <= n; ++i) {
        for (int j = i + 1; j <= n; ++j) {
            if (a[i] > a[j]) {
                int ri = find(find, i);
                int rj = find(find, j);
                if (ri == rj) {
                    if (col[i] == col[j]) {
                        cout << 0 << "\n";
                        return 0;
                    }
                } else {
                    p[ri] = rj;
                    col[ri] = col[i] ^ col[j] ^ 1;
                }
            }
        }
    }
    vector<int> c0(n + 1, 0), c1(n + 1, 0);
    for (int i = 1; i <= n; ++i) {
        int r = find(find, i);
        if (col[i] == 0) c0[r]++;
        else c1[r]++;
    }
    const int MOD = 1000000007;
    vector<int> dp(m + 1, 0);
    dp[0] = 1;
    for (int i = 1; i <= n; ++i) {
        if (p[i] != i) continue;
        int s0 = c0[i], s1 = c1[i];
        vector<int> ndp(m + 1, 0);
        for (int k = 0; k <= m; ++k) {
            if (!dp[k]) continue;
            if (k + s0 <= m) {
                ndp[k + s0] += dp[k];
                if (ndp[k + s0] >= MOD) ndp[k + s0] -= MOD;
            }
            if (k + s1 <= m) {
                ndp[k + s1] += dp[k];
                if (ndp[k + s1] >= MOD) ndp[k + s1] -= MOD;
            }
        }
        dp = move(ndp);
    }
    int ans = 0;
    for (int k = 0; k <= m; ++k) {
        if (n - k <= m) {
            ans += dp[k];
            if (ans >= MOD) ans -= MOD;
        }
    }
    cout << ans << "\n";
    return 0;
}