天梯赛L2题解（041-044）

L2-041 插松枝

模拟，感觉有点难分析，需要仔细读题分析

#include <bits/stdc++.h>
using namespace std;
#define int long long

using PII = pair<int, int>;
using AIT = array<int, 3>;

// 首先弄清楚，他优先是在小盒子里拿的，然后才会去推送器拿
// 一根插完只有两种可能，第一种就是没得拿了，第二种就是拿不到了

void ylh_() {
    int n, m, k;
    cin >> n >> m >> k;
    vector<int> a(n + 1);
    for (int i = 1; i <= n; ++i) {
        cin >> a[i];
    }
    queue<int> q; // 推送器
    stack<int> st; // 盒子
    vector<int> res; // 手上的松枝
    vector<vector<int>> ans; // 答案
    for (int i = 1; i <= n; ++i) {
        q.push(a[i]);
    }
    auto get = [&]() -> bool {
        if (res.size() == 0) {
            if (!st.empty()) {
                res.push_back(st.top());
                st.pop();
                return 1;
            }
            if (!q.empty()) {
                res.push_back(q.front());
                q.pop();
                return 1;
            }
        } else {
            if (!st.empty() && st.top() <= res.back()) {
                res.push_back(st.top());
                st.pop();
                return 1;
            }
            while (!q.empty() && q.front() > res.back() && st.size() < m) {
                st.push(q.front());
                q.pop();
            }
            if (!q.empty() && q.front() <= res.back()) {
                res.push_back(q.front());
                q.pop();
                return 1;
            }
        }
        return 0;
    };
    while (q.size() + st.size() > 0) {
        int f = get();
        // for (auto v : res) {
        //     cout << v << ' ';
        // }
        // cout << endl
        //      << f << endl;
        if (!f || res.size() == k) {
            ans.push_back(res);
            res.clear();
        }
    }
    if (res.size() != 0) {
        ans.push_back(res);
    }
    for (auto vec : ans) {
        for (int i = 0; i < vec.size(); ++i) {
            cout << vec[i] << " \n"[i == vec.size() - 1];
        }
    }
}

int32_t main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int T = 1;
    // cin >> T;
    while (T--) {
        ylh_();
    }
}

L2-042 老板的作息表

#include <bits/stdc++.h>
using namespace std;
#define int long long

using PII = pair<int, int>;

void ylh_() {
    auto trans_to = [&](string s) {
        int times = ((s[0] - '0') * 10 + (s[1] - '0')) * 3600 + ((s[3] - '0') * 10 + (s[4] - '0')) * 60 + ((s[6] - '0') * 10 + (s[7] - '0'));
        return times;
    };
    auto trans_back = [&](int x) {
        string s_hour, s_min, s_s;
        int hour = x / 3600, min = x / 60 % 60, s = x % 60;
        s_hour = ((hour < 10) ? "0" + to_string(hour) : to_string(hour));
        s_min = ((min < 10) ? "0" + to_string(min) : to_string(min));
        s_s = ((s < 10) ? "0" + to_string(s) : to_string(s));
        string res = s_hour + ":" + s_min + ":" + s_s;
        return res;
    };
    int n;
    cin >> n;
    vector<PII> a(n + 1);
    for (int i = 1; i <= n; ++i) {
        string s1, s2;
        cin >> s1 >> s2 >> s2;
        a[i] = { trans_to(s1), trans_to(s2) };
    }
    sort(a.begin() + 1, a.end());
    auto print = [&](int t1, int t2) -> void {
        cout << trans_back(t1) << " - " << trans_back(t2) << '\n';
    };
    if (a[1].first != 0) {
        print(0ll, a[1].first);
    }
    for (int i = 2; i <= n; ++i) {
        if (a[i].first != a[i - 1].second) {
            print(a[i - 1].second, a[i].first);
        }
    }
    const int ED = 23 * 3600 + 59 * 60 + 59;
    if (a[n].second != ED) {
        print(a[n].second, ED);
    }
}

int32_t main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int T = 1;
    // cin >> T;
    while (T--) {
        ylh_();
    }
}

L2-043 龙龙送外卖

怎么说呢，所有点到起点的经过边数都要走两遍，但是没说要回外卖站，所以减掉一个最深点的深度就行

#include <bits/stdc++.h>
using namespace std;
#define int long long

using PII = pair<int, int>;

void ylh_() {
    int n, m;
    cin >> n >> m;
    vector<vector<int>> g(n + 1), fg(n + 1);
    int rt = 0;
    for (int i = 1; i <= n; ++i) {
        int x;
        cin >> x;
        if (x == -1) {
            rt = i;
        } else {
            g[i].push_back(x);
            fg[x].push_back(i);
        }
    }
    vector<int> vis(n + 1), dep(n + 1);
    vis[rt] = 0;
    int Maxdep = 0;
    int Length = 0;
    auto dfs1 = [&](auto&& dfs, int u) -> void {
        for (auto v : fg[u]) {
            dep[v] = dep[u] + 1;
            dfs(dfs, v);
        }
    };
    dfs1(dfs1, rt);
    vis[rt] = 1;
    auto dfs = [&](auto&& dfs, int u, int step) -> void {
        // cout << '*' << u << endl;
        if (vis[u]) {
            Length += step;
            return;
        }
        vis[u] = 1;
        for (auto v : g[u]) {
            dfs(dfs, v, step + 1);
        }
    };
    for (int i = 1; i <= m; ++i) {
        int x;
        cin >> x;
        Maxdep = max(dep[x], Maxdep);
        dfs(dfs, x, 0);
        cout << Length * 2 - Maxdep << '\n';
    }
}

int32_t main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int T = 1;
    // cin >> T;
    while (T--) {
        ylh_();
    }
}

L2-044 大众情人

建立距离矩阵用floyd求最短路即可

#include <bits/stdc++.h>
using namespace std;
#define int long long
const int INF = 2e9;

void ylh_() {
    int n;
    cin >> n;
    vector<int> sex(n + 1);
    vector<vector<int>> dist(n + 1, vector<int>(n + 1, INF));
    for (int i = 1; i <= n; ++ i) {
        dist[i][i] = 0;
        char c;
        cin >> c;
        sex[i] = (c == 'M');
        int k;
        cin >> k;
        for (int j = 1; j <= k; ++ j) {
        	int p, x;
        	cin >> p >> c >> x;
        	dist[i][p] = x;
		}
    }
    for (int k = 1; k <= n; ++ k) {
    	for (int i = 1; i <= n; ++ i) {
    		for (int j = 1; j <= n; ++ j) {
    			dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j]);
			}
		}
	}
	int Min1 = INF, Min2 = INF;
	vector<int> ans1, ans2;
    for (int i = 1; i <= n; ++ i) {
    	int Max = 0;
    	for (int j = 1; j <= n; ++ j) {
    		if (i == j) continue;
    		if (sex[i] == sex[j]) continue;
    		Max = max(dist[j][i], Max);
		}
		if (sex[i] == 0) {
			if (Min1 > Max) {
				Min1 = Max;
				ans1.clear();
				ans1.push_back(i);
			} else if (Min1 == Max) {
				ans1.push_back(i);
			}
		} else {
			if (Min2 > Max) {
				Min2 = Max;
				ans2.clear();
				ans2.push_back(i);
			} else if (Min2 == Max) {
				ans2.push_back(i);
			}
		}
	}
	for (int i = 0; i < ans1.size(); ++ i) {
		if (i != 0) {
			cout << ' ';
		}
		cout << ans1[i];
	}
	cout << '\n';
	for (int i = 0; i < ans2.size(); ++ i) {
		if (i != 0) {
			cout << ' ';
		}
		cout << ans2[i];
	}
	cout << '\n';
}

int32_t main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int T = 1;
    // cin >> T;
    while (T--) {
        ylh_();
    }
}