2018-2019 ACM-ICPC, China Multi-Provincial Collegiate Programming Contest

A. Maximum Element In A Stack

按照題目模擬就好，棧內的最大值可以維護單調棧解決。

#include <bits/stdc++.h>

using namespace std;


using i64 = long long;
using ui32 = unsigned int;

ui32 SA, SB, SC;


ui32 rng61(){
    SA ^= SA << 16;
    SA ^= SA >> 5;
    SA ^= SA << 1;
    ui32 t = SA;
    SA = SB;
    SB = SC;
    SC ^= t ^ SA;
    return SC;
}

void solve() {
	int n, p, q, m;
	cin >> n >> p >> q >> m >> SA >> SB >> SC;
	stack<ui32> s;
	i64 res = 0;
	for(ui32 i = 1; i <= n ; i ++) {
		if(rng61() % (p + q) < p) {
			ui32 tmp = (rng61() % m + 1);
			if(s.empty()) s.push(tmp);
			else s.push(max(s.top(), tmp));
		} else if(not s.empty()) s.pop();
		if(not s.empty()) res ^= (i64)i * (i64)s.top();
	}
	cout << res << "\n";
}

int main() {
	ios::sync_with_stdio(false), cin.tie(nullptr);
	int TC;
	cin >> TC;
	for(int i = 1; i <= TC; i ++)
		cout << "Case #" << i << ": ", solve();
	return 0;
}

B. Rolling The Polygon

偏模板的題目，透過三個點算出夾角，然後根據半徑算一下弧長就好了。

#include<bits/stdc++.h>

using namespace std;

using i32 = int32_t;

#define int long long

using vi = vector<int>;
using pii = pair<int, int>;

const int inf = 1e9;

using db = long double;

struct Point {
    db x, y;

    Point(db x = 0, db y = 0) : x(x), y(y) {};
};

using Vec = Point;

const db pi = acosl(-1);

Vec operator+(Vec u, Vec v) { return Vec(u.x + v.x, u.y + v.y); }

Vec operator-(Vec u, Vec v) { return Vec(u.x - v.x, u.y - v.y); }

db operator*(Vec u, Vec v) { return u.x * v.x + u.y * v.y; }

db len(Vec v) { return sqrt(v.x * v.x + v.y * v.y); }

db cos_t(Vec u, Vec v) { return u * v / len(u) / len(v); }


using Points = vector<Point>;

ostream &operator<<(ostream &os, Point x) {
    return os << "(" << x.x << "," << x.y << ")";
}

void solve() {
    int n;
    cin >> n;
    Points ps(n);
    for (auto &[x, y]: ps) cin >> x >> y;
    Point po;
    cin >> po.x >> po.y;
    db res = 0;
    for (int i = 0, lst = n - 1, nxt = 1; i < n; i++, lst = (lst + 1) % n, nxt = (nxt + 1) % n) {
        Vec v1 = ps[nxt] - ps[i], v2 = ps[lst] - ps[i];
        db theta = acosl(cos_t(v1, v2));
        res += (pi - theta) * len(po - ps[i]);
    }
    cout << fixed << setprecision(3) << res << "\n";
    return;
}

i32 main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    int T = 1;
    cin >> T;
    for (int i = 1; i <= T; i++)
        cout << "Case #" << i << ": ", solve();
}

C. Caesar Cipher

#include <bits/stdc++.h>

using namespace std;


void solve() {
    int n, m;
    cin >> n >> m;
    string a, b, c;
    cin >> a >> b >> c;
    int d = (a.front() - b.front() + 26) % 26;
    for (auto i: c)
        cout << char((i - 'A' + d) % 26 + 'A');
    cout << "\n";
    return ;
}

signed main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    int T = 1;
    cin >> T;
    for( int i = 1 ; i <= T ; i ++ )
        cout << "Case #" << i << ": ", solve();
    return 0;
}

D. Take Your Seat

打表找找規律就好了

#include <bits/stdc++.h>

using namespace std;

#define ll long long

using vi = vector<int>;

void solve() {
    int n, m;
    cin >> n >> m;
    long double x = 1, y = 0;
    for (int i = 1; i <= m; i++)
        x += 1, y += 2;
    cout << fixed << setprecision(6) << (n == 1 ? 1.0 : 0.5) << " " << x / y << "\n";
    return;
}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int T;
    cin >> T;


    for (int i = 1; i <= T; i++) {
        cout << "Case #" << i << ": ";
        solve();
    }

    return 0;
}

F. Moving On

離線做，詢問按照\(w\)進行排序。然後每次加入一個點之後，把進行一次鬆弛操作。

#include<bits/stdc++.h>

using namespace std;

using i32 = int32_t;

#define int long long

using vi = vector<int>;
using pii = pair<int, int>;

const int inf = 1e9;

void solve() {
    int n, q;
    cin >> n >> q;
    vector<pii> a;
    for (int i = 1, x; i <= n; i++)
        cin >> x, a.emplace_back(x, i);
    sort(a.begin(), a.end());
    vector g(n + 1, vi(n + 1));
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++) cin >> g[i][j];
    vector<array<int, 4>> op(q);
    for (int id = 0; auto &[w, u, v, idx]: op)
        cin >> u >> v >> w, idx = id++;
    sort(op.begin(), op.end());
    vi res(q);
    for (int t = 0; const auto &[w, u, v, idx]: op) {
        while (t < n and a[t].first <= w) {
            int x = a[t].second;
            for (int i = 1; i <= n; i++)
                for (int j = 1; j < i; j++)
                    g[i][j] = g[j][i] = min(g[i][j], g[i][x] + g[x][j]);
            t++;
        }
        res[idx] = g[u][v];
    }
    for (auto &i: res) cout << i << "\n";
    return;
}

i32 main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    int T = 1;
    cin >> T;
    for (int i = 1; i <= T; i++)
        cout << "Case #" << i << ":\n", solve();
    return 0;
}

G. Factories

這本身是一個無根樹，所以先找一個不是葉子的點做根。

\(dp[i][j]\)表示\(i\)的子樹中選擇\(k\)個葉子點的最小代價，\(u\)是根節點，\(v\)是子節點，則轉移如下

\[dp[u][i] = dp[u][i-1] + dp[v][j] + w * (k - j) * j \]

其中\(w*(k-j)*j\)表示從\(v\)到\(u\)的邊產生的貢獻。

#include <bits/stdc++.h>

using namespace std;

#define ll long long

const ll inf = 1e18;

const int N = 1e5 + 6;

int n,k;
ll dp[N][110];
vector<pair<int,ll>> e[N];

void dfs(int u,int prv){
    for (auto [v,w] : e[u]) if (v != prv){
        dfs(v,u);
        for (int i = k;i >= 0;i--){
            for (int j = i;j >= 0;j--){
                dp[u][i] = min(dp[u][i],dp[u][i - j] + dp[v][j] + w * (k - j) * (j));
            }
        }
    }
}

void solve(){
    cin >> n >> k;

    for (int i = 1;i <= n;i++){
        e[i].clear();
    }

    for (int i = 1;i <= n - 1;i++){
        int u,v,w;cin >> u >> v >> w;
        e[u].push_back({v,w});
        e[v].push_back({u,w});
    }
    int root = -1;
    if (k == 1){
        cout << 0 << endl;
        return;
    }
    for (int i = 1;i <= n;i++){
        if (e[i].size() != 1){
            root = i;
            break;
        }
    }
    for (int i = 1;i <= n;i++){
        for (int j = 0;j <= k;j++){
            dp[i][j] = inf;
        }
    }
    for (int i = 1;i <= n;i++){
        dp[i][0] = 0;
    }
    for (int i = 1;i <= n;i++){
        if (e[i].size() == 1){
            dp[i][1] = 0;
        }
    }
    if (root == -1){
        int ans = 0;
        for (int i = 1;i <= n;i++){
            for (auto [v,w] : e[i]){
                ans += w;
            }
        }
        cout << ans / 2 << endl;
    }else{
        dfs(root,-1);
        cout << dp[root][k] << endl;
    }


}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int T;cin >> T;


    for (int i = 1;i <= T;i++){
        cout << "Case #" << i << ": ";
        solve();
    }

    return 0;
}

H. Fight Against Monsters

貪心做，先計算出擊敗一個怪物需要的次數\(cnt\)，然後按照\(\frac {stk} {cnt}\)從大到小逐個擊殺。

#include <bits/stdc++.h>

using namespace std;


using i32 = int32_t;
using i64 = long long;
#define int i64
using vi = vector<int>;

void solve() {
    int n;
    cin >> n;
    vector<tuple<double, int, int>> a;
    int sum = 0;
    for (int i = 1, hp, atk; i <= n; i++) {
        cin >> hp >> atk, sum += atk;
        int l = 1, r = hp, cnt = -1;
        while (l <= r) {
            int mid = (l + r) / 2;
            if (mid * (mid + 1) / 2 >= hp) cnt = mid, r = mid - 1;
            else l = mid + 1;
        }
        a.emplace_back(double(atk) / (double) cnt, atk, cnt);
    }
    sort(a.begin(), a.end(), greater<>());
    int res = 0;
    for (auto &[x, atk, cnt]: a) {
        res += cnt * sum;
        sum -= atk;
    }
    cout << res << "\n";
}

i32 main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    int TC;
    cin >> TC;
    for (int i = 1; i <= TC; i++)
        cout << "Case #" << i << ": ", solve();
    return 0;
}

K. Vertex Covers

我們可以把點分成兩部分，然後可以先計算左邊的部分，如果左邊有一個點不在集合中，則左邊所有與他相連的點都必須在集合內，否則不合法。

同理，在列舉右邊的集合。對於一個合法的集合，當右邊一個點不在集合中，則與這個點所有相連的左側的點必須在左側集合中，我們可以根據這個求出左邊集合的必選點，這左邊集合必須是必選點的超集。

我們可以用高維字首和，處理出左邊集合每個集合超集的和。

#include <bits/stdc++.h>

using namespace std;


using i32 = int32_t;

#define int long long

using vi = vector<int>;


void solve(){
	int n, m, p;
	cin >> n >> m >> p;

	vi val(n);
	for(auto &i : val) cin >> i;

	vector e(n ,vi(n));
	for(int u, v; m; m --) {
		cin >> u >> v, u --, v --;
		e[u][v] = e[v][u] = 1;
	}

	int n1 = n / 2, n2 = n - n1;
	vi f(1<<n1);
	for(int i = 0; i < (1<<n1); i ++ ){
		auto tmp = 1;
		for(int j = 0; j < n1 and tmp; j ++){
			if(i & (1<<j)) tmp = tmp * val[j] % p;
			else {
				for(int k = 0; k < n1 and tmp; k ++)
					if(not (i & (1<<k)) and e[j][k] ) // k 也不在集合中，且 j,k 之間有邊 
						tmp = 0;
			}
		}
		f[i] = tmp;
	}

	// SOS DP, 列舉超集
	for(int j = 0; j < n1; j ++)
		for(int i = 0; i < (1<<n1); i ++)
			if(not(i &(1<<j))) f[i] = (f[i] + f[i ^ (1<<j)]) % p;

	int res = 0;
	for(int i = 0; i < (1<<n2); i ++){
		int tmp = 1;
		for( int j = 0; j < n2 and tmp; j ++){
			if(i & (1<<j)) tmp = tmp * val[n1 + j] % p;
			else {
				for(int k = 0; k < n2 and tmp; k ++)
					if(not (i & (1<<k)) and e[n1 + j][n1 + k])
						tmp = 0;
			}
		}
		if(tmp == 0) continue;
		int need = 0;
		for(int j = 0; j < n2; j ++){
			if(i & (1<<j)) continue;
			for(int k = 0; k < n1; k ++) 
				if(e[n1 + j][k]) // j 沒被選 且 j,k 之間存在邊，則 k 必選
					need |= (1<<k);
		}
		res = (res + tmp * f[need] % p) % p;
	}
	cout << res << "\n";
	return;
}

i32 main(){
	ios::sync_with_stdio(false), cin.tie(nullptr);
	int T;
	cin >> T;
	for(int i = 1; i <= T; i ++)
		cout << "Case #" << i << ": ", solve();
	return 0;
}