Educational Codeforces Round 62 (Rated for Div. 2) 2019年3月23日

A.

Example

Input

9
1 3 3 6 7 6 8 8 9
Output

4

 

#include <iostream>

using namespace std;

int main () {
    int n;
    cin >> n;
    int a[10010];
    for(int i = 0;i < n;i++){
        cin >> a[i];
    }
    int day = 0;
    int t = 0;
    for(int i = 1;i <= n;i++){
        t = max(a[i - 1],t);
        if(i == t)
            day++;
    }
    cout << day << endl;
    return 0;
}

B

Example

Input

3
2
<>
3
><<
1
>
Output

1
0
0

 

 

// 本题问的是先删除后再按照上述判断 进而看是否能构成
#include <bits/stdc++.h>

using namespace std;

int main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin >> t;
    while (t--) {
        int n;
        cin >> n;
        string s;
        cin >> s;
        int l = 0;
        int r = n - 1;
        while(l < n && s[l] == '<')
            l++;
        cout << "l :" << l << endl;
        while(r >= 0 && s[r] == '>')
            r--;
        cout << "r " << r << endl;
        cout << min(l, n - r - 1) << endl;
        // 从左边开始‘<’的连续个数与从右边开始‘>’的连续个数 两者的最小值即为答案
        // 因为除去这个边界之后的下一个数一定与它相反 通过它就可以实现转化
    }
}

 

C

Examples

Input

4 3
4 7
15 1
3 6
6 8
Output

78
Input

5 3
12 31
112 4
100 100
13 55
55 50

Output
10000

#include<bits/stdc++.h>
using namespace std;
const int N = 3e5+1;
long long n, k, x, y, s, A;

pair<int,int>p[N];

priority_queue<int>q;
// 优先队列就是大顶堆，队头元素最大
int main(){
	cin >> n >> k;
	for (int i = 1; i <= n; i++) {			// 从第一个开始进行存入
		cin >> x >> y;
		p[i] = make_pair(y, x);
	}
	sort(p + 1, p + n + 1);						// pair 默认对first升序，当first相同时对second升序（就是对beauty values进行排序）
	for(int i = n; i >= 1; i--) {
		s += p[i].second;				// 将 lengths 从大到小相加
		q.push(-p[i].second);		// 将length的相反数放入队列中
		if (q.size() > k) {				// 当压入的数目超过最大个数时
			s += q.top();					// s加上最上面压入的元素（正值最小的）
			q.pop();						// 弹出最上面的
		}								
		A = max(A, p[i].first * s);			// p[i].first一定是该状态下的最大值
	}
	cout << A << endl;
	return 0;
}

 

D

// 一个多变形最多能被分成 n - 2个不重合的三角形
#include <bits/stdc++.h>

using namespace std;

int main () {
	int n;
	cin >> n;
	int sum = 0;
	for (int i = 2;i < n; i++) {
		sum += i * (i + 1);
	}
	cout << sum << endl;
}

 

待补！！！

E

 

 

#include <bits/stdc++.h>

using namespace std;

#define MOD 998244353

int n, k, x[2], at;
long long dp[2][2];
int ta;

int main() {
	scanf("%d%d", &n, &k);
	for (int i = 0; i < n; i++) {
		at = i & 1;
		scanf("%d", &ta);
		if (ta == -1) {
			if (x[at] == 0) {
				x[at] = 1;
				dp[at][0] = dp[at][1] = 1;
			} else {
				long long tmp = dp[at][0];
				dp[at][0] = (dp[at][0] * (k - 2) + dp[at][1]) % MOD;
				dp[at][1] = tmp * (k - 1) % MOD;
			}
		} else {
			if (x[at] == 0)
				dp[at][1] = 1;
			else if (ta == x[at])
				dp[at][1] = dp[at][0] * (k - 1) % MOD;
			else
				dp[at][1] = (dp[at][0] * (k - 2) + dp[at][1]) % MOD;
			dp[at][0] = 0;
			x[at] = ta;
		}
	}
	for (int i = 0; i < 2; i++)
		dp[i][0] = (dp[i][0] * (k - 1) + dp[i][1]) % MOD;
	printf("%lld\n", dp[0][0] * dp[1][0] % MOD);
}

 

F

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define f(i, x, n) for (int i = x; i < (int)(n); ++i)

int const N = 300000;
int q;
vector<pair<int, int> > s[N << 2];
map<pair<int, int>, int> edtm;
vector<pair<int, int> > p;
ll an[N + 1];

void adsg(pair<int, int> e, int x, int y, int l = 1, int r = q, int id = 1){
	if (l > y || r < x)return;
	if (l >= x && r <= y) { s[id].push_back(e); return; }
	int m = l + r >> 1, a = id << 1, b = a | 1;
	adsg(e, x, y, l, m, a);
	adsg(e, x, y, m + 1, r, b);
}

int pa[N << 1 | 1];
pair<int, int> sz[N << 1 | 1];
stack<pair<int, int> > st;
ll can;

int P(int p) { while (pa[p])p = pa[p]; return p; }
void adan(pair<int, int> p) { can += (ll)p.first * p.second; }
void sban(pair<int, int> p) { can -= (ll)p.first * p.second; }

void fn(int l = 1, int r = q, int id = 1){
	int undc = 0;
	for (auto &t: s[id]){
		int a = P(t.first), b = P(t.second);
		if (a == b)continue;
		if (sz[a].first + sz[a].second < sz[b].first + sz[b].second)swap(a, b);
		pa[b] = a;
		sban(sz[a]);
		sban(sz[b]);
		sz[a].first += sz[b].first;
		sz[a].second += sz[b].second;
		adan(sz[a]);
		st.push(make_pair(a, b));
		++undc;
	}
	if (l == r)an[l] += can;
	else {
		int m = l + r >> 1, a = id << 1, b = a | 1;
		fn(l, m, a);
		fn(m + 1, r, b);
	}
	while (undc--){
		int a = st.top().first, b = st.top().second;
		st.pop();
		sban(sz[a]);
		sz[a].first -= sz[b].first;
		sz[a].second -= sz[b].second;
		adan(sz[a]);
		adan(sz[b]);
		pa[b] = 0;
	}
}

 

G

 

 

 

 

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define f(i, x, n) for (int i = x; i < (int)(n); ++i)

ll const inf = 1e18;
int const N = 300000, B = 19;
ll wsd[N + 1], w[N + 1][B][2][2];
int pa[N + 1][B], dp[N + 1];

struct E{
	int u;
	ll w[2];
	E(int a, ll b, ll c):u(a) { w[0] = b, w[1] = c; }
};

vector<E> tr[N + 1];

struct D{
	ll w[2];
	D(int v) { w[0] = w[1] = 0; w[v & 1] = inf; }
	void operator +=(ll c[][2]){
		ll nw0 = min(w[0] + c[0][0], w[1] + c[1][0]);
		w[1] = min(w[0] + c[0][1], w[1] + c[1][1]);
		w[0] = nw0;
	}
};

void J(ll a[][2], ll b[][2], ll c[][2]){
	f(i, 0, 2)f(j, 0, 2)c[i][j] = inf;
	f(i, 0, 2)f(j, 0, 2)f(k, 0, 2)c[i][j] = min(c[i][j], a[i][k] + b[k][j]);
}

void pl1(int v = 1, int p = 0){
	for (E &t: tr[v])if (t.u != p)pl1(t.u, v);
	for (E &t: tr[v])if (t.u != p)wsd[v] = min(wsd[v], wsd[t.u] + t.w[0] + t.w[1]);
}

void pl2(int v = 1, int p = 0) { for (E &t: tr[v])if (t.u != p)wsd[t.u] = min(wsd[t.u], wsd[v] + t.w[0] + t.w[1]), pl2(t.u, v); }

void pl3(int v = 1, int p = 0){
	pa[v][0] = p;
	dp[v] = dp[p] + 1;
	int k = 1, z;
	while (k < B && (z = pa[v][k - 1])){
		pa[v][k] = pa[z][k - 1];
		J(w[v][k - 1], w[z][k - 1], w[v][k]);
		++k;
	}
	for (E &t: tr[v]){
		if (t.u == p)continue;
		w[t.u][0][0][0] = min(t.w[0], t.w[1] + wsd[v] + wsd[t.u]);
		w[t.u][0][0][1] = min(t.w[0] + wsd[v], wsd[t.u] + t.w[1]);
		w[t.u][0][1][0] = min(t.w[1] + wsd[v], wsd[t.u] + t.w[0]);
		w[t.u][0][1][1] = min(t.w[1], t.w[0] + wsd[v] + wsd[t.u]);
		pl3(t.u, v);
	}
}

ll wan[2][2];

ll go(int a, int b){
	D ca(a), cb(b);
	a = a + 1 >> 1, b = b + 1 >> 1;
	if (dp[a] < dp[b])swap(a, b), swap(cb, ca);
	int d = dp[a] - dp[b], j = 0;
	while (d){
		if (d & 1)ca += w[a][j], a = pa[a][j];
		++j;
		d >>= 1;
	}
	if (a != b){
		for (int i = B - 1; i >= 0; --i)if (pa[a][i] != pa[b][i])ca += w[a][i], cb += w[b][i], a = pa[a][i], b = pa[b][i];
		ca += w[a][0], cb += w[b][0];
		a = pa[a][0];
	}
	ll an = inf;
	f(i, 0, 2)f(j, 0, 2)an = min(an, ca.w[i] + cb.w[j] + (i != j ? wsd[a] : 0));
	return an;
}

int main(){
	int n;
	scanf("%d", &n);
	f(i, 1, n + 1)scanf("%lld", wsd + i);
	f(i, 1, n){
		int a, b;
		ll w1, w2;
		scanf("%d%d%lld%lld", &a, &b, &w1, &w2);
		tr[a].push_back(E(b, w1, w2));
		tr[b].push_back(E(a, w1, w2));
	}
	pl1();
	pl2();
	pl3();
	int q;
	scanf("%d", &q);
	while (q--){
		int a, b;
		scanf("%d%d", &a, &b);
		printf("%lld\n", go(a, b));
	}
}