Toyota Programming Contest 2023#4（AtCoder Beginner Contest 311）

A - First ABC (atcoder.jp)

记录一下\(ABC\)是否都出现过了然后输出下标

#include <bits/stdc++.h>
#define int long long

using namespace std;

signed main() {

    ios::sync_with_stdio(false);cin.tie(nullptr);

    int n;
    string s;
    cin >> n >> s;
    bitset<3> v;
    for(int i = 0;i < n;i ++){
        v[s[i] - 'A'] = 1;
        if(v[0] && v[1] && v[2]){
            cout << i + 1<< endl;
            break;
        }
    }

    return 0;
}

B - Vacation Together (atcoder.jp)

数据范围小暴力寻找即可

#include <bits/stdc++.h>
#define int long long

using namespace std;

signed main() {

    ios::sync_with_stdio(false);cin.tie(nullptr);

    int n,d;
    cin >> n >> d;
    vector<string> g(n);
    for (int i = 0; i < n; ++i) {
        cin >> g[i];
    }

    int ans = 0;
    for(int i = 0;i < d ;i++){
        for(int j = d;j >= i; j--){
            bool f = true;
            for(auto k : g){
                if(k.substr(i,j - i + 1) != string(j - i + 1,'o'))
                    f = false;
            }
            if(f) ans = max(ans, j - i + 1);
        }
    }
    cout << ans << endl;

    return 0;
}

还是jiangly的思路巧妙啊,用一个初始化全为\(true\)的bool数组就好了, 每次去跑一遍字符串,将对应位置\(x\)更新为\(false\),最后找出连续最长的\(true\)就好了,这个比暴力找快了7,8倍

#include <bits/stdc++.h>

using i64 = long long;

int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    
    int N, D;
    std::cin >> N >> D;
    
    std::vector<bool> ok(D, true);
    for (int i = 0; i < N; i++) {
        std::string s;
        std::cin >> s;
        for (int j = 0; j < D; j++) {
            if (s[j] == 'x') {
                ok[j] = false;
            }
        }
    }
    
    int ans = 0;
    int len = 0;
    for (auto x : ok) {
        len = x ? 1 + len : 0;
        ans = std::max(ans, len);
    }
    std::cout << ans << "\n";
    
    return 0;
}

C - Find it! (atcoder.jp)

题意就是找出一个环

哎,我还是传统dfs的

#include <bits/stdc++.h>
#define int long long

using namespace std;

const int N = 2e5 + 10;

signed main() {

    ios::sync_with_stdio(false);cin.tie(nullptr);

    int n;
    cin >> n;
    vector<int> g[n + 1];
    vector<int> fa(n + 1);
    iota(fa.begin(),fa.end(),0);
    for(int i = 1,a ;i <= n;i ++){
        cin >> a;
        g[i].push_back(a);
    }

    bool ok = false;
    vector<int> ans;
    bitset<N> vis;
    int len = 1;
    function<void(int,int)> dfs = [&](int u,int v) -> void {
        for(auto i : g[u]){
            if(vis[i]){
                int p = 0;
                for(auto j : ans){
                    if(j != i)
                        p++;
                    else
                        break;
                }
                cout << len - p << endl;
                for(int j = p;j < ans.size();j ++ ){
                    cout << ans[j] << ' ';
                }
                exit(0);
            }
            if(i == v) continue;
            len++;
            ans.push_back(i);
            vis[i] = 1;
            dfs(i,u);
            len--;
            vis[i] = 0;
            ans.pop_back();
        }
    };

    for(int i = 1;i <= n;i ++){
        vis.reset();
        len = 1;
        vis[i] = 1;
        ans.push_back(i);
        dfs(i,i);
        ans.pop_back();
    }

    return 0;
}

每日佩服jiangly(

他的思路很巧妙啊(~~或许是我太傻了~~,用一个链表的思路就解决了

#include <bits/stdc++.h>

using i64 = long long;

int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    
    int N;
    std::cin >> N;
    
    std::vector<int> A(N);
    for (int i = 0; i < N; i++) {
        std::cin >> A[i];
        A[i]--;
    }
    
    std::vector<bool> vis(N);
    int i = 0;
    while (!vis[i]) {
        vis[i] = true;
        i = A[i];
    }
    int j = i;
    std::vector<int> ans;
    do {
        ans.push_back(j);
        j = A[j];
    } while (j != i);
    std::cout << ans.size() << "\n";
    for (int i = 0; i < ans.size(); i++) {
        std::cout << ans[i] + 1 << " \n"[i == ans.size() - 1];
    }
    
    return 0;
}

D - Grid Ice Floor (atcoder.jp)

借鉴jiangly,其实赛时我感觉咱的思路也和他很接近了,但是一直没调出来,看了jiangly的才发现他是一个vis数组判断是否走过,一个pass数组记录走过的路径,而我是一个vis数组将两个功能一起合体了(~~或许就是这里错了吧~~

#include <bits/stdc++.h>
#define int long long

using namespace std;

const int N = 2e5 + 10;

signed main() {

    ios::sync_with_stdio(false);cin.tie(nullptr);

    int n,m;
    cin >> n >> m;
    vector<string> g(n);
    vector<bitset<320>> vis(320),pass(320);
    for(auto &i : g) cin >> i;

    int u[] = {0,0,-1,1};
    int v[] = {1,-1,0,0};
    queue<pair<int,int>> Q;
    Q.emplace(1,1);
    pass[1][1] = vis[1][1] = true;
    while(Q.size()){
        auto [x,y] = Q.front();
        Q.pop();

        for(int i = 0;i < 4;i ++){
            int dx = x;
            int dy = y;

            while(g[dx][dy] == '.'){
                pass[dx][dy] = true;
                dx += u[i];
                dy += v[i];
            }
            dx -= u[i];
            dy -= v[i];
            if(!vis[dx][dy]){
                vis[dx][dy] = true;
                Q.emplace(dx,dy);
            }
        }
    }

    int ans = 0;
    for(int i = 1;i < n;i ++)
        for(int j = 1;j < m;j ++)
            ans += pass[i][j];
    cout << ans << endl;

    return 0;
}