模拟赛#1 | USACO19 DEC Bronze

题目链接:
A https://www.luogu.com.cn/problem/P5831
B https://www.luogu.com.cn/problem/P5832
C https://www.luogu.com.cn/problem/P5832

---

A

分析: 范围很小, 直接暴力枚举即可, 每一对奶牛枚举看看是不是一致, 考察循环程序设计, 时间复杂度\(O(n^2k)\)

代码:

/*
Author: SJ
*/
#include<bits/stdc++.h>
const int N = 1e2 + 10;
using ll = long long;
using ull = unsigned long long;

int k, n, re[N][N];
int main() {
	std::ios::sync_with_stdio(false);
	std::cin.tie(nullptr);

	std::cin >> k >> n;
	for (int i = 1; i <= k; i++) {
		for (int j = 1; j <= n; j++) {
			int x;
			std::cin >> x;
			re[x][i] = j;
		}
	}
	ll ans = 0;
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) {
			if (i != j) {
				bool flag = 1;
				for (int p = 1; p <= k; p++) {
					if (re[i][p] < re[j][p]) {
						flag = 0;
						break;
					}
				}
				if (flag) ans++;
			}
		}
	}
	std::cout << ans;
	return 0;
}

---

B

分析: 继续暴力, 考察string相关函数, 字符串操作, 时间复杂度\(O(n^3)\)

代码

/*
Author: SJ
*/
#include<bits/stdc++.h>
const int N = 1e5 + 10;
using ll = long long;
using ull = unsigned long long;

int n;
std::string s1;
std::vector<std::string> v1;
int main() {
	std::ios::sync_with_stdio(false);
	std::cin.tie(nullptr);

	std::cin >> n >> s1;
	for (int k = n; k >= 1; k--) {
		for (int i = 0; i + k <= s1.size(); i++) {
			std::string s3 = s1.substr(i, k);
			// std::cout << s3 << ' ' << k << "\n";
			for (auto i : v1) {
				if (s3 == i) {
					std::cout << k + 1;
					return 0;
				}
			}
			v1.push_back(s3);
		}
	}
	return 0;
}

C

分析: 用next_permutation可以直接秒的, 莫名其妙挖掘了一堆没用的性质, 然后上了个贪心(强行加大代码难度bushi

代码:

/*
Author: SJ
*/
#include<bits/stdc++.h>
const int N = 1e5 + 10;
using ll = long long;
using ull = unsigned long long;

std::string s[10] = {
	"Zeka", "Bessie", "Buttercup", "Belinda", "Beatrice", "Bella", "Blue", "Betsy", "Sue"
};
std::vector<int> g[10];
bool vis[N];
std::queue<int> q;
int query() {
	int head = 0;
	for (int i = 1; i <= 8; i++) {
		if (!vis[i] && s[i] < s[head] && g[i].size() <= 1) {
			head = i;
		}
	}
	return head;
}
bool cmp(int a, int b) {
	return s[a] < s[b];
}
void dfs(int x) {
	std::cout << s[x] << "\n";
	vis[x] = 1;
	for (auto i : g[x]) {
		if (!vis[i]) {
			dfs(i);
		}
	}
}
int main() {
	std::ios::sync_with_stdio(false);
	std::cin.tie(nullptr);

	int n;
	std::cin >> n;
	for (int i = 1; i <= n; i++) {
		std::string s1, s2;
		std::cin >> s1;
		int mark = 5;
		while (mark--) std::cin >> s2;
		// std::cout << s1 << ' ' << s2 << "\n";

		int mark1, mark2;
		for (int i = 1; i <= 8; i++) {
			if (s[i] == s1) mark1 = i;
			else if (s[i] == s2) mark2 = i;
		}
		g[mark1].push_back(mark2);
		g[mark2].push_back(mark1);
	}
	for (int i = 1; i <= 8; i++) std::sort(g[i].begin(), g[i].end(), cmp);
	for (int i = 1; i <= 8; i++) {
		int head = query();
		if (!head) break;
		dfs(head);
	}
	return 0;
}

这个贪心很显然是对的

用next_permutation的代码:

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