Codeforces Round #699 (Div. 2)

Codeforces Round #699 (Div. 2)

A Space Navigation

int main() {
    IOS;
    for (cin >> _; _; --_) {
        int x, y; cin >> x >> y; string s; cin >> s;
        for (auto c : s)
            if (x > 0 && c == 'R') --x;
            else if (x < 0 && c == 'L') ++x;
            else if (y > 0 && c == 'U') --y;
            else if (y < 0 && c == 'D') ++y;
        cout << (x || y ? "NO
" : "YES
");
    }
    return 0;
}

B New Colony

按照题意模拟石头的下落过程, 反正数据小

int main() {
    IOS;
    for (cin >> _; _; --_) {
        cin >> n >> k; VI h(n); m = -1;
        for (auto &i : h) cin >> i;
        rep (i, 1, k) {
            rep (j, 1, n - 1)
                if (h[j] > h[j - 1]) { ++h[j - 1], m = j; break; }
                else if (j == n - 1) { m = -1; break; }
            if (m == -1) break;
        }
        cout << (m > -1 ? m : -1) << '
';
    }
    return 0;
}

C Fence Painting

弄个两个集合, 一个是必须要涂一次颜色i的集合, 一种是颜色i随便涂的集合

把不能图的颜色放在将涂这两个集合内之前的模板就行

int a[N], b[N], c[N], p[N], d[N];
 
int main() {
    IOS;
    for (cin >> _; _; --_) {
        cin >> n >> m; vector<VI> col(n + 1);
        rep(i, 1, n) cin >> a[i], p[i] = 0;
        rep(i, 1, n) {
            cin >> b[i];
            if (a[i] == b[i]) p[b[i]] = i;
            else col[b[i]].pb(i);
        }
        rep(i, 1, m) {
            cin >> c[i];
            if (!col[c[i]].empty()) {
                while (!col[0].empty()) d[col[0].back()] = col[c[i]].back(), col[0].pop_back();
                d[i] = col[c[i]].back(); col[c[i]].pop_back(); p[c[i]] = d[i];
            } else if (p[c[i]]) {
                while (!col[0].empty()) d[col[0].back()] = p[c[i]], col[0].pop_back();
                d[i] = p[c[i]];
            }
            else col[0].pb(i);
        }
        bool f = 1;
        for (auto& i : col) if (!i.empty()) f = 0;
        if (!f) { cout << "NO
"; continue; }
        cout << "YES
"; rep(i, 1, m) cout << d[i] << ' '; cout << '
';
    }
    return 0;
}

D AB Graph

奇数直接在两个点之前反复跳即可

偶数要找3个点, d[a][b] = d[b][c]

其他无解

int main() {
    IOS;
    for (cin >> _; _; --_) {
        cin >> n >> k; bool f = 0;
        rep (i, 1, n) cin >> d[i] + 1;
        rep (i, 1, n) rep (j, i + 1, n) if (d[i][j] == d[j][i] && !f) {
            cout << "YES
"; cout << i << ' ';
            rep (t, 1, k) cout << (t & 1 ? j : i) << ' ';
            cout << '
'; f = 1;
        }
        if (f) continue;
        if (k & 1) {
            cout << "YES
"; cout << 1 << ' ';
            rep (i, 1, k) cout << (i & 1 ? 2 : 1) << ' ';
            cout << '
';
        }
        else if (n < 3) { cout << "NO
"; continue; }
        else {
            cout << "YES
"; int a = 0, b = 1, c = 2;
            while (d[a + 1][b + 1] != d[b + 1][c + 1]) a = (a + 1) %  3, b = (b + 1) % 3, c = (c + 1) % 3;
            //既然到了这里, 那么就不存在 d[a][b] == d[b][a], 那么 d[a][b] = d[b][c], 那么必然 (d[b][a] = d[c][b]) != (d[a][b] = d[b][c])
            vector<int> t = {a + 1, b + 1, c + 1, b + 1};
            rep (i, 0, k) cout << t[(i + (k % 4 == 0)) % 4] << ' '; cout << '
';
        }
    }
    return 0;
}