Atcoder Beginner Contest 373(A-D)

AtCoder Beginner Contest 373

A - September

模拟，直接写就行

代码

void solve()
{
    int ans = 0;
    for (int i = 1; i <= 12; i++)
    {
        std::string s;
        std::cin >> s;
        if (s.size() == i) ans++;
    }
    std::cout << ans << endl;
}

B - 1D Keyboard

模拟，直接写就行

代码

void solve()
{
    int ans = 0;
    charmap mp;
    for (int i = 1; i <= 26; i++)
    {
        char ch;
        std::cin >> ch;
        mp[ch] = i;
    }
    for (int i = 1; i <= 25; i++)
    {
        char tmp = (char)('A' + i);
        char tmp2 = (char)('A' + i - 1);
        ans += abs(mp[tmp] - mp[tmp2]);
    }
    std::cout << ans << endl;
}

C - Max Ai+Bj

模拟，直接写就行

代码

void solve()
{
    int a = -inf, b = -inf;
    int tmp = 0;
    int n = 0;
    std::cin >> n;
    for (int i = 1; i <= n; i++)
    {
        std::cin >> tmp;
        a = std::max(a, tmp);
    }
    for (int i = 1; i <= n; i++)
    {
        std::cin >> tmp;
        b = std::max(b, tmp);
    }
    std::cout << a + b << endl;
}

D - Hidden Weights

思路

1. 因为一定会有解，所以按照\(ans[end] = ans[start] + weight\)的方式把图遍历一遍就行

2. 虽然题目里面是单向图，但是一定要建成双向图————因为有的节点没有入度，遍历到它时还没有更新也会\(vis[x] = 1\)，这样以后就无法被更新了

代码

// #pragma GCC optimize(1)
// #pragma GCC optimize(2)
// #pragma GCC optimize(3,"Ofast","inline")
#include <bits/stdc++.h>
#define endl '\n'
#define int long long
using lli = long long;
using ull = unsigned long long;
typedef std::pair<int, int> pair;
typedef std::unordered_map<int, int> intmap;
typedef std::unordered_map<char, int> charmap;
const int inf = 0x7f7f7f7f;
bool vis[400010];
int ans[400010];
struct Edge
{
    int start;
    int terminus;
    int weight;
    int nxt;
}edge[400010];
int head[400010];
int cnt = 0;
void AddEdge(int s, int t, int w)
{
    edge[++cnt].start = s;
    edge[cnt].terminus = t;
    edge[cnt].weight = w;
    edge[cnt].nxt = head[s];
    head[s] = cnt;
}

void dfs(int x)
{
    if (vis[x]) return;
    vis[x] = 1;
    for (int i = head[x]; i != -1; i = edge[i].nxt)
    {
        if (vis[edge[i].terminus]) continue;
        ans[edge[i].terminus] = ans[edge[i].start] + edge[i].weight;
        dfs(edge[i].terminus);
    }
}  
void solve()
{
    std::fill(head, head + 400010, -1);
    int n = 0, m = 0;
    std::cin >> n >> m;
    int u = 0, v = 0, w = 0;
    for (int i = 1; i <= m; i++)
    {
        std::cin >> u >> v >> w;
        AddEdge(u, v, w);
        AddEdge(v, u, -w);
    }
    for (int i = 1; i <= n; i++)
    {
        if (!vis[i])
        {
            dfs(i);
        }
    }
    for (int i = 1; i <= n; i++)
    {
        std::cout << ans[i] << ' ';
    }
}

signed main()
{
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr); std::cout.tie(nullptr);
    //freopen("out.txt", "w", stdout);
    int t = 1;
    //std::cin >> t;
    while(t--)
    {
        solve();
    }
    return 0;
}