牛客小白月赛29

进攻

排序贪心

ll a[N];
PLL b[N];

int main() {
	IOS; cin >> n >> m;
	rep (i, 1, n) cin >> a[i];
	rep (i, 1, m) cin >> b[i].se;
	rep (i, 1, m) cin >> b[i].fi;
	sort(b + 1, b + 1 + m, greater<PLL>());
	sort(a + 1, a + 1 + n, greater<ll>());
	ll ans = 0;
	for (int i = 1, j = 1; i <= n && j <= m; ++i) {
		while (j <= m && b[j].se >= a[i]) ++j;
		if (j > m) break;
		if (b[j].fi > 0) ans += b[j].fi;
        else break;
	}
	cout << ans;
	return 0;
}

二进制

拆位处理, 计算当前输入的数 第 i 位(二进制) 为 1/0 输出是什么

然后判断什么样的操作会使答案一样, 比如对于 第i为

与 x = (1<<20) - 1, 这样不会影响答案及 a & x = a

或 y = 0, 及 a | y = a

异或 z = 0, 及 a ^ z = 0

1/0 -> 1/1 必定是 x ^= 1 << i

1/0 -> 1/0 没改变

1/0 -> 0/1 z ^= 1 << i

1/0 -> 0/0 x ^= 1 << i

三次运算即可, 及 &x, |y, ^z

int main() {
	IOS; cin >> n;
	int a = (1 << 20) - 1, b = 0;
	rep (i, 1, n) {
		int op, x; cin >> op >> x;
		if (op == 1) a &= x, b &= x;
		else if (op == 2) a |= x, b |= x;
		else a ^= x, b ^= x;
	}
	int x = (1 << 20) - 1, y = 0, z = 0;
	rep (i, 0, 19)
		if (!(a >> i & 1))
			if (b >> i & 1) z ^= 1 << i;
			else x ^= 1 << i;
		else if (b >> i & 1) y ^= 1 << i;
	cout << "3
3 " << z << "
1 " << x << "
2 " << y << '
';
	return 0;
}

种树

往下遍历就行

ll w[N], c[N], dep[N], ans;
PII a[N];

void dfs(int x) {
    if (a[x].fi) dep[a[x].fi] = dep[x] + 1, dfs(a[x].fi);
    if (a[x].se) dep[a[x].se] = dep[x] + 1, dfs(a[x].se);
    if (a[x].fi == 0 && a[x].se == 0 && dep[x] <= m) umax(ans, w[x]);
}

int main() {
	IOS; cin >> n; m = ((n - 1) / 2 + 1) / 2;
    rep (i, 1, n) cin >> a[i].fi >> a[i].se;
    rep (i, 1, n) cin >> w[i];
    ans = -2e9;
    dfs(1); cout << ans;
	return 0;
}

积木

构造吧, 就是四个同色拼成一个大的正方形就行

int main() {
	IOS; cin >> n;
    if (n & 1) { cout << -1; return 0; }
    int f = 1;
    for (int i = 1; i <= n; i += 2) {
        for (int j = 1, g = f; j <= n; j += 2, g ^= 1)
                a[i][j] = a[i][j + 1] = a[i + 1][j] = a[i + 1][j + 1] = g;
        f ^= 1;
    }
    f = 0;
    rep (k, 1, n) {
        rep (i, 1, n) {
            rep (j, 1, n) cout << (a[i][j] ^ f) << ' ';
            cout << '
';
        }
        f ^= 1;
    }
	return 0;
}

考试

贪心

int main() {
	IOS; cin >> n >> k;
    rep (i, 1, n) cin >> a[i];
    rep (i, 1, n) {
        cin >> _;
        if (_ == a[i]) ++m;
        else if (k) --k, ++m;
    }
    m -= k; cout << m;
	return 0;
}

项链

链表而已

int ne[N][2];
 
int main() {
    IOS; cin >> n >> m;
    rep (i, 1, n + 1) ne[i - 1][k ^ 1] = i, ne[i][k] = i - 1;
    rep (i, 1, m) {
        int op; cin >> op;
        if (op == 1) {
            int x, y; cin >> x >> y;
            ne[ne[x][k]][k ^ 1] = ne[x][k ^ 1];
            ne[ne[x][k ^ 1]][k] = ne[x][k];
            ne[x][k] = y; ne[x][k ^ 1] = ne[y][k ^ 1];
            ne[ne[y][k ^ 1]][k] = x;
            ne[y][k ^ 1] = x;
        } else if (op == 2) {
            int x, y; cin >> x >> y;
            ne[ne[x][k]][k ^ 1] = ne[x][k ^ 1];
            ne[ne[x][k ^ 1]][k] = ne[x][k];
            ne[x][k] = ne[y][k]; ne[x][k ^ 1] = y;
            ne[ne[y][k]][k ^ 1] = x;
            ne[y][k] = x;
        }
        else if (op == 3) k ^= 1;
        else {
            int i = 1;
            for (i = 1; i != n + 1 && i; i = ne[i][k ^ 1]) cout << i << ' ';
            for (i = (i ? ne[0][k ^ 1] : ne[n + 1][k ^ 1]); i != 1; i = ne[i][k ^ 1]) cout << i << ' ';
            cout << '
';
        }
    }
    return 0;
}

涂色

签到

int main() {
      IOS; cin >> n;
      cout << n + 1;
      return 0;
}

园

签到

int main() {
    IOS;
    for (cin >> _; _; --_) {
        cin >> x >> y >> r >> a >> b >> c;
        if (r < c) swap(x, a), swap(y, b), swap(r, c);
        ll d = sqr(x - a) + sqr(y - b);
        long double dd = sqrt(d);
        if (dd + c < r) cout << "NO
";
        else if (dd + c == r) cout << "YES
";
        else if (d > sqr(r + c)) cout << "NO
";
        else cout << "YES
";
    }
    return 0;
}

修改

给你的序列a是任意的, 我们就人为添加一个 点(n+1) 且 a[n+1] = 0

每次操作 l, r, 可以使得 a[l] 和 a[r + 1] 变得相同

最终目标是所有点都可以和 a[n+1] 相同变成 0

这不就是最小生成树吗?

int find(int x) {
    return x == f[x] ? x : f[x] = find(f[x]);
}

int main() {
    IOS; cin >> n >> m; ++n;
    rep (i, 1, n) f[i] = i;
    rep (i ,1, m) cin >> a[i].se.fi >> a[i].se.se >> a[i].fi, ++a[i].se.se;
    sort(a  + 1, a + 1 + m);
    ll ans = 0, cnt = 0;
    rep (i, 1, m) {
        int x = find(a[i].se.fi), y = find(a[i].se.se);
        if (x != y) f[x] = y, ans += a[i].fi, ++cnt;
    }
    cout << (cnt == n - 1 ? ans : -1);
    return 0;
}

克隆

发现是连通图， 剩下的不用说了吧

VI h[N];
int st[N << 1], top;
bool v[N];

void dfs(int x) {
    v[x] = 1; st[++top] = x;
    for (auto y : h[x]) if (!v[y]) dfs(y), st[++top] = x;
}

int main() {
	IOS; cin >> n >> m >> k;
    rep (i, 1, m) {
        int u, v; cin >> u >> v;
        h[u].pb(v); h[v].pb(u);
    }
    int a = ((n << 1) - 1) / k + 1;
    dfs(1); cout << "YES
"; --top;
    for (int c = 1; c <= top; --k) {
        int d = min(a, top - c + 1); 
        cout << d << ' ';
        rep (i, 1, d) cout << st[c++] << ' ';
        cout << '
';
    }
    while (k--) cout << "0
";
	return 0;
}