AtCoder Beginner Contest 051

A - Haiku

直接模拟。

#include <bits/stdc++.h>

using namespace std;
using i64 = long long;

int main() {
	ios::sync_with_stdio(false), cin.tie(nullptr);
	string s;
	cin >> s;
	string a, b, c;
	a = s.substr(0, 5);
	b = s.substr(6, 7);
	c = s.substr(14);
	cout << a << " " << b << " " << c;
	return 0;
}

B - Sum of Three Integers

暴力做的话是三重循环会超时，可以枚举前两个数，然后判断第三个数是否合法即可，时间复杂度为 \(O(n^2)\)。

#include <bits/stdc++.h>

using namespace std;
using i64 = long long;

int main() {
	ios::sync_with_stdio(false), cin.tie(nullptr);
	int K, S;
	cin >> K >> S;
	int ans = 0;
	for (int i = 0; i <= K; i++) {
		for (int j = 0; j <= K; j++) {
			int k = S - i - j;
			if (k >= 0 && k <= K) ans++;
		}
	}
	cout << ans;
	return 0;
}

C - Back and Forth

本题看着吓人，其实只需要找到一条互相不重复的路线即可，不妨就以样例 \(1\) 为例。画出图形，然后模拟即可。

#include <bits/stdc++.h>

using namespace std;
using i64 = long long;

int main() {
	ios::sync_with_stdio(false), cin.tie(nullptr);
	int x, y, u, v;
	cin >> x >> y >> u >> v;
	for (int i = y; i < v; i++) cout << 'U';
	for (int i = x; i < u; i++) cout << 'R';
	for (int i = v; i > y; i--) cout << 'D';
	for (int i = u; i > x; i--) cout << 'L';
	cout << 'L';
	for (int i = y; i < v + 1; i++) cout << 'U';
	for (int i = x - 1; i < u; i++) cout << 'R';
	cout << "DR";
	for (int i = v; i > y - 1; i--) cout << 'D';
	for (int i = u + 1; i > x; i--) cout << 'L';
	cout << 'U';
	return 0;
}

D - Candidates of No Shortest Paths

\(\rm Floyd\) 算法，之后会补。

#include<bits/stdc++.h>

using i64 = long long;

void DAOQI() {
    int n, m;
    std::cin >> n >> m;
    std::vector d(n + 1, std::vector(n + 1, (i64) 1e9));
    std::vector<std::array<int, 3>> doc;
    for (int i = 1; i <= m; i++) {
        int u, v, w;
        std::cin >> u >> v >> w;
        d[u][v] = w;
        d[v][u] = w;
        doc.push_back({u, v, w});
    }

    for (int k = 1; k <= n; k++) {
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= n; j++) {
                d[i][j] = std::min(d[i][j], d[i][k] + d[k][j]);
            }
        }
    }

    int ans = 0;
    for (auto [u, v, w]: doc) {
        if (d[u][v] < w) ans++;
    }
    std::cout << ans << "\n";
}

signed main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    int T = 1;
    //std::cin >> T;
    while (T--) DAOQI();
    return 0;
}

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

int main() {
    
    int n, m;
    cin >> n >> m;
    
    vector dp(n, vector<int>(n, 1e9));
    for(int i = 0; i < n; i++) dp[i][i] = 0;
    
    vector<int> a(m), b(m), c(m);
    for(int i = 0; i < m; i++) {
        cin >> a[i] >> b[i] >> c[i];
        a[i]--; b[i]--;
        dp[a[i]][b[i]] = c[i];
        dp[b[i]][a[i]] = c[i];
    }
    
    for(int k = 0; k < n; k++) {
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < n; j++) {
                dp[i][j] = min(dp[i][j], dp[i][k] + dp[k][j]);
            }
        }    
    }
    
    int cnt = 0;
    for(int i = 0; i < m; i++) {
        if(dp[a[i]][b[i]] < c[i]) cnt++;
    }

    cout << cnt << endl;
    
    return 0;
}