【开源社区openEuler实践】F. Turtle Mission: Robot and the Earthquake

题目

题解：

按题解思路的代码：

#include <bits/stdc++.h>

using i64 = long long;

void solve() {
    int n, m;
    std::cin >> n >> m;
    
    std::vector a(n, std::vector<int>(m));
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            std::cin >> a[i][j];
        }
    }
    
    std::queue<std::pair<int, int>> q;
    std::vector dis(n, std::vector(m, -1));
    q.emplace(0, 0);
    dis[0][0] = 0;
    while (!q.empty()) {
        auto [x, y] = q.front();
        q.pop();
        
        if (y < m - 1 && a[(x + 1) % n][y + 1] == 0 && dis[(x + 1) % n][y + 1] == -1) {
            dis[(x + 1) % n][y + 1] = dis[x][y] + 1;
            q.emplace((x + 1) % n, y + 1);
        }
        if (a[(x + 1) % n][y] == 0 && a[(x + 2) % n][y] == 0 && dis[(x + 2) % n][y] == -1) {
            dis[(x + 2) % n][y] = dis[x][y] + 1;
            q.emplace((x + 2) % n, y);
        }
    }
    
    int ans = -1;
    for (int i = 0; i < n; i++) {
        int res = dis[i][m - 1];
        if (res != -1) {
            if (res % n != (i + 1) % n) {
                res += (i + 1 - res % n + n) % n;
            }
            if (ans == -1 || ans > res) {
                ans = res;
            }
        }
    }
    std::cout << ans << "\n";
}

int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    
    int t;
    std::cin >> t;
    
    while (t--) {
        solve();
    }
    
    return 0;
}

另一种方法：

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define pb push_back
#define fi first
#define se second
#define lson p << 1
#define rson p << 1 | 1
#define ll long long
#define pii pair<int, int>
#define ld long double
const int maxn = 1e6 + 5, inf = 1e18, maxm = 4e4 + 5, base = 37;
const int N = 2e6;
const int mod = 1e9 + 7;
// const int mod = 998244353;
// const __int128 mod = 212370440130137957LL;
int n, m;

//long long ? maxn ? n? m?
void solve(){
    ll res = 0;
    int k;
    cin >> n >> m;
	// vector<int> a(n + 1);
	// for(int i = 1; i <= n; i++){
	// 	cin >> a[i];
	// }
    vector<vector<int>> a(n + 1, vector<int>(m + 1)), dis(n + 1, vector<int>(m + 1, inf));
    for(int i = 0; i < n; i++){
        for(int j = 0; j < m; j++){
            cin >> a[i][j];
        }
    }
    queue<pii> q;
    dis[0][0] = 0;
    q.push({0, 0});
    while(!q.empty()){
        auto [x, y] = q.front();
        q.pop();
        int nx = (x + 1) % n, ny = y;
        if(dis[nx][ny] == inf && a[(x + 1 + dis[x][y]) % n][y] == 0 && a[(x + 2 + dis[x][y]) % n][y] == 0){
            dis[nx][ny] = dis[x][y] + 1;
            q.push({nx, ny});
        }
        nx = x, ny = y + 1;
        if(ny < m && dis[nx][ny] == inf && a[(x + 1 + dis[x][y]) % n][y + 1] == 0){
            dis[nx][ny] = dis[x][y] + 1;
            q.push({nx, ny});
        }
    }
    res = inf;
    for(int i = 0; i < n; i++){
        res = min({res, dis[i][m - 1] + min(n - 1 - i, i + 1)});
    }
    if(res == inf) res = -1;
    cout << res << '\n';
}
    
signed main(){
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout << fixed << setprecision(9);
    
    int T = 1;
    cin >> T;
    while (T--){
        solve();
    }
    return 0;
}