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📂SMU2024训练记录
🔖BFS/DFS动态规划离线数论
2024-03-07 23:45阅读: 10评论: 0推荐: 0

2024天梯选拔赛(一)

2024天梯选拔赛(一)

A 私人笑声

#include <bits/stdc++.h>
#define debug(a) cout<<#a<<"="<<a<<'\n';
 
using namespace std;
using i64 = long long;
 
typedef pair<i64, i64> PII;
 
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
 
    string str;
    getline(cin, str);
    for (int i = 0; i < str.size(); i ++) {
        cout << str[i] ;
        if (str[i] == '.')
            cout << "xixixixi.";
        }
 
    return 0;
}

B 孵化小鸡

数据小,dfs

#include <bits/stdc++.h>
#define debug(a) cout<<#a<<"="<<a<<'\n';
 
using namespace std;
using i64 = long long;
 
typedef pair<i64, i64> PII;
 
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
 
    int n, M;
    cin >> n >> M;
    vector<int> a(n), b(n), m(n), l(m), r(m), k(m), p(m);
    for (int i = 0; i < n; i ++)
        cin >> a[i] >> b[i] >> m[i];
    for (int i = 0; i < M; i ++)
        cin >> l[i] >> r[i] >> k[i] >> p[i];
 
    vector<bool> vis(M);
    vector<int> lr(200);
    i64 ans = INT_MAX;
    auto dfs = [&](auto self, int num, i64 res) {
        if (num == M) {
            bool ok = true;
            for (int i = 0; i < n; i ++) {
                if (!ok) break;
                for (int j = a[i]; j <= b[i]; j ++) {
                    if (lr[j] < m[i]) {
                        ok = false;
                        break;
                    }
                }
            }
            if (ok) ans = min(ans, res);
            return ;
        }
 
        self(self, num + 1, res);
        for (int i = l[num]; i <= r[num]; i ++)
            lr[i] += k[num];
        self(self, num + 1, res + p[num]);
        for (int i = l[num]; i <= r[num]; i ++)
            lr[i] -= k[num];
    };
 
    dfs(dfs, 0, 0);
    cout << ans << '\n';
 
    return 0;
}

C 可怕的冻雨

考虑离线;

将落脚点按光滑程度排序,以及雪地靴按防滑程度排序;

预处理出能够直接到达的落脚点记录并存储两两点之间的距离;

遍历雪地靴,判断雪地靴能否在原有的落脚点上新增落脚点,能够新增则找到新增点的邻近点,更新原有的距离,最后判断雪地靴的行走距离能否跨过最远的冰层;

防滑程度小的雪地靴能到达的落脚点,防滑度大的可延续其之前的路线;

#include <bits/stdc++.h>
#define debug(a) cout<<#a<<"="<<a<<'\n';
 
using namespace std;
using i64 = long long;
 
typedef pair<i64, i64> PII;
 
int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
 
	int n, m;
	cin >> n >> m;
	vector<PII> A;
	for (int i = 1, x; i <= n; i ++) {
		cin >> x;
		A.emplace_back(x, i);
	}
 
	vector<array<int, 3>> B;
	for (int i = 0, k, s; i < m; i ++) {
		cin >> k >> s;
		B.push_back({k, s, i});
	}
 
	sort(A.begin(), A.end());
	sort(B.begin(), B.end());
 
	set<int> loc;
	multiset<int> dis;
	int index = 0;
	while (index < A.size() && !A[index].first)
		loc.insert(A[index].second), index ++;
	for (auto it = loc.begin(); next(it) != loc.end(); it = next(it))
		dis.insert(*next(it) - *it);
 
	vector<int> ans(m);
	for (auto [k, s, id] : B) {
		while (index < A.size() && A[index].first <= k) {
			int i = A[index].second;
			index ++;
 
			auto t = loc.upper_bound(i);
			int r = *t, l = *(prev(t));
			loc.insert(i);
			dis.erase(dis.find(r - l));
			dis.insert(i - l);
			dis.insert(r - i);
		}
		ans[id] = (s >= *dis.rbegin());
	}
 
	for (auto i : ans)
		cout << i << '\n';
 
	return 0;
}

D 划分田地（easy）

枚举矩形,将在矩形内的点加入进去;

#include <bits/stdc++.h>
#define debug(a) cout<<#a<<"="<<a<<'\n';
 
using namespace std;
using i64 = long long;
 
typedef pair<i64, i64> PII;
 
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
 
    int n;
    cin >> n;
    vector<PII> tu(n);
    for (auto &[x, y] : tu)
        cin >> x >> y;
 
    set<vector<int>> ans;
    for (int i = 0; i <= 50; i ++) {
        for (int j = 0; j <= 50; j ++) {
            for (int k = i; k <= 50; k ++) {
                for (int p = j; p <= 50; p ++) {
                    vector<int> ve;
                    int num = 0;
                    for (auto [x, y] : tu) {
                        num ++;
                        if (x >= i && x <= k && y >= j && y <= p)
                            ve.push_back(num);
                    }
                    ans.insert(ve);
                }
            }
        }
    }
 
    cout << ans.size() << '\n';
 
    return 0;
}

E 划分田地（hard）

对于一个矩形,如果它的上边扫描到上边界这个矩形中有\(x\)个点,下边扫描到下边界的这个矩形中有\(y\)个点,那么这个矩形可以向上向下拓展边界从而可以得到\((x + 1) \times (y + 1)\)个包含不同点的矩形(向左向右也是一样);

本题中,我采用的是上下拓展;

将坐标二维离散后,运用二维前缀和计算上矩形和下矩形中分别有多少个点,从而累加计算结果, 最后加上\(n+1\)是\(n\)个点和一个空集;

#include <bits/stdc++.h>
#define debug(a) cout<<#a<<"="<<a<<'\n';
 
using namespace std;
using i64 = long long;
 
typedef pair<i64, i64> PII;
 
struct Two_D_Discrete {
	int n, tot1 = 1, tot2 = 1;
	vector<vector<int>> mp;
	vector<int> x, y, nx, ny;
	vector<pair<i64, i64>> a;
	vector<PII> New;
 
	Two_D_Discrete (int _n, vector<pair<i64, i64>>& _a): n(_n), a(_a) {
		x.resize(n), y.resize(n);
		nx.resize(n * 2 + 5), ny.resize(n * 2 + 5);
		vector<vector<int>>(n * 2 + 5, vector<int>(n * 2 + 5)).swap(mp);
		for (int i = 0; i < n; i ++) {
			x[i] = a[i].first;
			y[i] = a[i].second;
		}
	}
 
	void work() {
		//排序
		sort(x.begin(), x.end());
		sort(y.begin(), y.end());
		// 去重 并得到有多少个点
		int len1 = unique(x.begin(), x.end()) - x.begin();
		int len2 = unique(y.begin(), y.end()) - y.begin();
		// 离散化 x 轴
		for (int i = 0; i < len1; i++) {
			if (i && x[i] != x[i - 1] + 1)
				nx[tot1++] = x[i] - 1, nx[tot1++] = x[i];
			else
				nx[tot1++] = x[i];
		}
		// 离散化 y 轴
		for (int i = 0; i < len2; i++) {
			if (i && y[i] != y[i - 1] + 1)
				ny[tot2++] = y[i] - 1, ny[tot2++] = y[i];
			else
				ny[tot2++] = y[i];
		}
		//映射关系将需离散的点放入离散图中
		for (int i = 0; i < n; i++) {
			int newx = lower_bound(nx.begin(), nx.begin() + tot1, a[i].first) - nx.begin();
			int newy = lower_bound(ny.begin(), ny.begin() + tot2, a[i].second) - ny.begin();
			mp[newx][newy] = 1;
			// cout << "(" << newx << ',' << newy << ")\n";
			New.emplace_back(newx, newy);
		}
	}
};
 
int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
 
	int n;
	cin >> n;
	vector<PII> a(n);
	for (auto &[x, y] : a) {
		cin >> x >> y;
		x ++, y ++;
	}
 
	Two_D_Discrete _2D(n, a);
	_2D.work();
 
	int Ke = n * 2 + 5;
	vector sum(Ke, vector<int>(Ke));
 
	for (int i = 1; i < Ke; i ++)
		for (int j = 1; j < Ke; j ++) 
			sum[i][j] = sum[i - 1][j] + sum[i][j - 1] - sum[i - 1][j - 1] + _2D.mp[i][j];
 
	auto calc = [&](int a, int b, int c, int d) {
		return sum[c][d] - sum[a - 1][d] - sum[c][b - 1] + sum[a - 1][b - 1];
	};
 
	i64 ans = 0;
	for (int i = 0; i < n; i ++) {
		for (int j = i + 1; j < n; j ++) {
			auto [x1, y1] = _2D.New[i];
			auto [x2, y2] = _2D.New[j];
			if (x1 > x2) swap(x1, x2);
			if (y1 > y2) swap(y1, y2);
 
			int num1 = calc(1, y1, x1 - 1, y2);
			int num2 = calc(x2 + 1, y1, Ke - 1, y2);
			
			ans += (num1 + 1) * (num2 + 1);
		}
	}
 
	cout << ans + n + 1 << '\n';
 
	return 0;
}

F 加一余二

用set存储连续相同子串区间,multiset存储其区间长度;

对于操作的下标\(x\),要去判断子串区间中所在的位置,进行分裂与合并;

#include <bits/stdc++.h>
#define debug(a) cout<<#a<<"="<<a<<'\n';
 
using namespace std;
using i64 = long long;
 
typedef pair<i64, i64> PII;
 
int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
 
	string s;
	int m;
	cin >> s >> m;
	multiset<int> ans;
	set<PII> res;
 
	for (int i = 1, j; i <= s.size(); i = j + 1) {
		j = i;
		while (j < s.size() && s[j] == s[i - 1]) j ++;
		res.insert(PII(i, j));
		ans.insert(j - i + 1);
	}
 
	auto len = [](PII x) {
		return x.second - x.first + 1;
	};
 
	auto Find = [&](int x) {
		auto t = res.lower_bound(PII(x, -1));
		if (t == res.end() || (*t).first > x) t = prev(t);
		return *t;
	};
 
	while (m --) {
		int x;
		cin >> x;
 
		PII t = Find(x);
		res.erase(res.find(t));
		ans.erase(ans.find(len(t)));
		if (x != 1) {
			if (t.first == x) {
				PII p = Find(x - 1);
				res.erase(res.find(p));
				ans.erase(ans.find(len(p)));
				t.first = p.first;
			} else {
				PII p = {t.first, x - 1};
				t.first = x;
				res.insert(p);
				ans.insert(len(p));
			}
		}
		if (x != s.size()) {
			if (t.second == x) {
				PII p = Find(x + 1);
				res.erase(res.find(p));
				ans.erase(ans.find(len(p)));
				t.second = p.second;
			} else {
				PII p = {x + 1, t.second};
				t.second = x;
				res.insert(p);
				ans.insert(len(p));
			}
		}
		res.insert(t);
		ans.insert(len(t));
 
		cout << *ans.rbegin() << ' ';
	}
 
	return 0;
}

G 相加余三（easy）

本题分别模拟三种情况然后取最大值即可。

#include <bits/stdc++.h>
#define debug(a) cout<<#a<<"="<<a<<'\n';
 
using namespace std;
using i64 = long long;
 
typedef pair<i64, i64> PII;
 
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
 
    int n;
    cin >> n;
    vector<int> a(n);
    for (auto &i : a) cin >> i;
 
    i64 ans1 = 0 , ans2 = 0 , ans3 = 0 ;
    for (int i = 0; i < n; i += 2) {
        ans1 += ((a[i] + a[i + 1]) % 3);
    }
    for (int i = n; i >= 0; i -= 2) {
        ans2 += ((a[i] + a[i - 1]) % 3);
    }
    for (int i = 0, j = n - 1; i < j; i ++, j --) {
        ans3 += ((a[i] + a[j]) % 3);
    }
    cout << max({ans1, ans2, ans3}) << '\n';
 
    return 0;
}

H 相加余三（hard）

考虑区间\(dp\);

从大区间往小区间转移,分奇偶讨论一下是因为每次去掉偶数个数,所以对于一个区间,到左右边界的值为奇数时是不合法的;

\[dp[i+2][j]=max(dp[i+2][j],dp[i][j]+(a[i]+a[i+1])%3) \]

\[dp[i][j-2]=max(dp[i][j-2],dp[i][j]+(a[j-1]+a[j])%3); \]

\[dp[i + 1][j - 1] = max(dp[i + 1][j - 1], dp[i][j] + (a[i] + a[j]) % 3); \]

#include <bits/stdc++.h>
#define debug(a) cout<<#a<<"="<<a<<'\n';
 
using namespace std;
using i64 = long long;
 
typedef pair<i64, i64> PII;
 
int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
 
	int n;
	cin >> n;
	vector<int> a(n + 1);
	for (int i = 1; i <= n; i ++)
		cin >> a[i];
 
	vector dp(n + 1, vector<int>(n + 1));
	int ans = 0;
	for (int i = 1; i <= n; i ++) {
		for (int j = n; j > i; j --) {
 
			if (!((j - i) & 1) && n % 2 == 0) continue;
			if (((j - i) & 1) && (n & 1)) continue;
 
			if (i + 1 <= n) ans = max(ans, dp[i][i + 1] + (a[i] + a[i + 1]) % 3);
			if (j >= i + 2) {
				dp[i + 2][j] = max(dp[i + 2][j], dp[i][j] + (a[i] + a[i + 1]) % 3);
				dp[i][j - 2] = max(dp[i][j - 2], dp[i][j] + (a[j - 1] + a[j]) % 3);
				dp[i + 1][j - 1] = max(dp[i + 1][j - 1], dp[i][j] + (a[i] + a[j]) % 3);
			}
		}
	}
 
	for (int i = 1; i <= n; i ++)
		for (int j = 1; j <= n; j ++)
			ans = max(ans, dp[i][j]);
 
	cout << ans << '\n';
 
	return 0;
}

从小区间往大区间转移,预处理小区间的值

\[dp[i][j] = max(dp[i][j], dp[i + 2][j] + (a[i] + a[i + 1]) % 3); \]

\[dp[i][j] = max(dp[i][j], dp[i][j - 2] + (a[j] + a[j - 1]) % 3); \]

\[dp[i][j] = max(dp[i][j], dp[i + 1][j - 1] + (a[i] + a[j]) % 3); \]

#include <bits/stdc++.h>
#define debug(a) cout<<#a<<"="<<a<<'\n';
 
using namespace std;
using i64 = long long;
 
typedef pair<i64, i64> PII;
 
int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
 
	int n;
	cin >> n;
	vector<int> a(n + 1);
	for (int i = 1; i <= n; i ++)
		cin >> a[i];
 
	vector dp(n + 1, vector<int>(n + 1));
 
	int ans = 0;
	for (int len = 2; len <= n; len ++) {
		for (int i = 1, j = len; j <= n; j ++, i ++) {
			if (len == 2)
				dp[i][j] = (a[i] + a[j]) % 3;
			else {
				dp[i][j] = max(dp[i][j], dp[i + 2][j] + (a[i] + a[i + 1]) % 3);
				dp[i][j] = max(dp[i][j], dp[i][j - 2] + (a[j] + a[j - 1]) % 3);
				dp[i][j] = max(dp[i][j], dp[i + 1][j - 1] + (a[i] + a[j]) % 3);
			}
 
		}
	}
 
	cout << dp[1][n] << '\n';
 
	return 0;
}

I 找除数

一个正整数\(n\)可以表示为\(n=p_1^{x_1} \times p_2^{x_2} \times p_3^{x_3} \times p_4^{x_4} \dots\)(其中\(p_i\)为质数)

则\(n\)的除数的数量$=(x_1 + 1) \times (x_2 + 1) \times (x_3 + 1) \times (x_4 + 1) \times \dots $

预处理出10000以内的质数;

然后将\(n\)按质数分解,按照上述公式计算即可;

#include <bits/stdc++.h>
#define debug(a) cout<<#a<<"="<<a<<'\n';
 
using namespace std;
using i64 = long long;
 
typedef pair<i64, i64> PII;
 
vector<int> prime, isnp(10005);
 
void solve() {
 
	int n;
	cin >> n;
	int ans = 1;
	for (auto k : prime) {
		if (n % k == 0) {
			int cnt = 1;
			while (n % k == 0) n /= k, cnt ++;
			ans *= cnt;
		}
	}
	if (n != 1) ans *= 2;
 
	cout << ans << '\n';
 
}
 
int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
 
	for (int i = 2; i <= 10000; i ++) {
		if (!isnp[i]) prime.push_back(i);
		for (auto k : prime) {
			if (k * i > 10000) break;
			isnp[i * k] = 1;
			if (i % k == 0) break;
		}
	}
	
	int T;
	cin >> T;
 
	while (T --)
		solve();
 
	return 0;
}

J 最后都是0

\(j\)为整数\(n\)上的某一位数字.

\[f[n] = inf \]

\[f[n]=min(f[n],f[n-j] + 1) \]

#include <bits/stdc++.h>
#define debug(a) cout<<#a<<"="<<a<<'\n';
 
using namespace std;
using i64 = long long;
 
typedef pair<i64, i64> PII;
 
int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
 
	int n;
	cin >> n;
	vector<int> dp(n + 1, INT_MAX);
	dp[0] = 0;
 
	for (int i = 1; i <= n; i ++) {
		string s = to_string(i);
		for (auto j : s) {
			dp[i] = min(dp[i], dp[i - (j - '0')] + 1);
		}
	}
 
	cout << dp[n] << '\n';
 
	return 0;
}

K 第五人格，启动！

包括起点和终点一共五个点,计算出两两点之间的距离,用全排列表示出所有路线,然后取最短时间

#include <bits/stdc++.h>
#define debug(a) cout<<#a<<"="<<a<<'\n';
 
using namespace std;
using i64 = long long;
 
typedef pair<i64, i64> PII;
 
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
 
    int n, m;
    cin >> n >> m;
    int u[] = {1, -1, 0, 0}, v[] = {0, 0, 1, -1};
    vector<string> mp(n);
    for (auto &i : mp) cin >> i;
    int sx, sy, ex, ey;
    vector<int> t(3);
    vector<PII> xy(3);
    cin >> sx >> sy ;
    sx --, sy --;
    for (auto &[x, y] : xy) {
        cin >> x >> y;
        x--, y--;
    }
    cin >> ex >> ey;
    ex--, ey--;
    for (auto &i : t) cin >> i;
    vector dis(5, vector<i64>(5, 0));
    for (int i = 0; i < 3; i ++) {
        auto [x, y] = xy[i];
        dis[0][i + 1] = abs(x - sx) + abs(y - sy);
    }
 
    for (int i = 0; i < 3; i ++)
        for (int j = 0; j < 3; j ++)
            dis[i + 1][j + 1] = (abs(xy[i].first - xy[j].first) + abs(xy[i].second - xy[j].second));
 
 
    auto bfs = [&](int x, int y, int edx, int edy) -> i64{
        vector<bitset<110>> vis(m);
        queue<pair<PII, i64>> Q;
        Q.push({{x, y}, 0});
        vis[x][y] = 1;
        int res = 0;
        while (Q.size()) {
            auto [axy, w] = Q.front();
            auto [ax, ay] = axy;
            Q.pop();
            if (ax == edx && ay == edy) {
                res = w;
                break;
            }
            for (int i = 0; i < 4; i ++) {
                int dx = u[i] + ax;
                int dy = v[i] + ay;
                if (dx >= 0 && dx < n && dy >= 0 && dy < m && !vis[dx][dy] && mp[dx][dy] != '#') {
                    Q.push({{dx, dy}, w + 1});
                    vis[dx][dy] = 1;
                }
            }
        }
        return res;
    };
 
    for (int i = 0; i < 3; i ++) {
        auto [x, y] = xy[i];
        dis[i + 1][4] = bfs(x, y, ex, ey);
    }
 
    i64 ans = LLONG_MAX;
    vector<vector<int>> st;
    vector<int> sh{1, 2, 3};
    do {
        vector<int> jk{0};
        for (int i = 0; i < 3; i ++) {
            jk.push_back(sh[i]);
        }
        jk.push_back(4);
        st.push_back(jk);
    } while (next_permutation(sh.begin(), sh.end()));
 
    for (auto v : st) {
        i64 res = 0, p = 1;
        for (int i = 1; i < 5; i ++) {
            res += dis[v[i - 1]][v[i]] * p;
            if (p != 4) p += t[v[i] - 1];
        }
        ans = min(ans, res);
    }
 
    cout << ans << '\n';
 
    return 0;
}

L 加纳~

已知视频一共完整播放了 k / n 遍，然后我们需要判断一下剩下的时间是否大于 m 秒即可。

#include <bits/stdc++.h>
#define debug(a) cout<<#a<<"="<<a<<'\n';
 
using namespace std;
using i64 = long long;
 
typedef pair<i64, i64> PII;
 
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
 
    i64 n, m, k;
    cin >> n >> m >> k;
    cout << k / n + (k % n >= m) << '\n';
 
    return 0;
}

本文作者：Ke_scholar

本文链接：https://www.cnblogs.com/Kescholar/p/18060068

版权声明：本作品采用知识共享署名-非商业性使用-禁止演绎 2.5 中国大陆许可协议进行许可。

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