Codeforces Round #642 (Div. 3)(A,B,C,D,E,F)

A. Most Unstable Array

思路:
分情況討論
代碼：

#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
typedef vector<int> vi;
typedef queue<int> qi;
typedef long long ll;
typedef pair<int, int> pii;
typedef double db;

int main()
{
	int t; scanf("%d", &t);
	while (t --)
	{
		int n, m; cin >> n >> m;
		if(n == 1) cout << 0 << endl;
		else if(n == 2) cout << m << endl;
		else cout << 2 * m << endl;
		
	}

	return 0;
}

B. Two Arrays And Swaps

思路:
貪心
取a數組的前k小的值和b數組的前k大的值進行交換就行了
代碼：

#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
typedef vector<int> vi;
typedef queue<int> qi;
typedef long long ll;
typedef pair<int, int> pii;
typedef double db;
const int N = 35;
int a[N], b[N];

int main()
{
	int t; cin >> t;
	while (t --)
	{
		int n, k; cin >> n >> k;
		for (int i = 1; i <= n; ++ i) cin >> a[i];
		for (int i = 1; i <= n; ++ i) cin >> b[i];
		sort (a + 1, a + n + 1);
		sort (b + 1, b + n + 1);
		reverse(b + 1, b + n + 1);
		
		int ans = 0;
		for (int i = 1; i <= n; ++ i)
		{
			if(k && b[i] > a[i]) ans += b[i], k --;
			else ans += a[i];
		}
		cout << ans << endl;
	}

	return 0;
}

C. Board Moves

思路：
肯定是移到中間的代價最小，
可以在紙上寫出每個位置的代價，就可以看出規律，類似於水的波紋那種，一圈一圈的，
代碼：

#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
typedef vector<int> vi;
typedef queue<int> qi;
typedef long long ll;
typedef pair<int, int> pii;
typedef double db;

int main()
{
	int t; cin >> t;
	while (t --)
	{
		int n; cin >> n;
		ll ans = 0;
		for (int i = 1; i <= (n - 1) / 2; ++ i) ans += 1LL * 8 * i * i;
		cout << ans << endl;
	}
	
	return 0;
}

D. Constructing the Array

思路:
直接按題目要求用優先隊列模擬即可，時間複雜度$O(nlogn)$
代碼：

#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
typedef vector<int> vi;
typedef queue<int> qi;
typedef long long ll;
typedef pair<int, int> pii;
typedef double db;
const int N = 2e5 + 5; 
struct seg
{
	int l, r;
	bool operator<(const seg &a)const 
	{
		if(r - l != a. r - a.l) return (r - l + 1) < (a.r - a.l + 1); 
		else return l > a.l;
	}
};
int ans[N];
priority_queue<seg> q;

int main()
{
	int t; cin >> t;
	while (t --)
	{
		int n; cin >> n;
		while (q.size()) q.pop();
		q.push(seg{1, n});
		int p = 1;
		while (q.size() && p <= n)
		{
			int l = q.top().l, r = q.top().r; q.pop();
			int pos;
			if((r - l + 1) % 2 == 1)
			{
				pos = (l + r) / 2;
				ans[pos] = p ++;
			}
			else
			{
				pos = (l + r - 1) / 2;
				ans[pos] = p ++;
			}
			if(l < pos) q.push(seg{l, pos - 1});
			if(pos < r) q.push(seg{pos + 1, r});
		}
		for (int i = 1; i <= n; ++ i) cout << ans[i] << " ";
		cout << endl;
	}

	return 0;
}

E. K-periodic Garland

這題比賽的時候沒開出來，我還是太菜了

思路:
dp[i,0]表示第i棧燈滅的情況下並且[1-i]的燈是合法的最小花費，
dp[i,1]表示第i棧燈亮的情況下並且[1-i]的燈是合法的最小花費

代碼：

#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
typedef vector<int> vi;
typedef queue<int> qi;
typedef long long ll;
typedef pair<int, int> pii;
typedef double db;
const int N = 1e6 + 5;
char s[N];
int dp[N][2], sum[N];
int main()
{
	int t; scanf("%d", &t);
	while (t --)
	{
		int n, k; scanf("%d%d", &n, &k);
		scanf("%s", s + 1);
		for (int i = 1; i <= n; ++ i) sum[i] = sum[i - 1] + (s[i] == '1');
		for (int i = 1; i <= n; ++ i)
		{
			int p = max(0, i - k);
			dp[i][0] = min(dp[i - 1][0], dp[i - 1][1]) + (s[i] == '1');
			dp[i][1] = min(sum[i - 1], dp[p][1] + sum[i - 1] - sum[p]) + (s[i] == '0');
//			cout << i << " " << dp[i][0] << " " << dp[i][1] << endl;
		} 
		printf("%d\n", min(dp[n][0], dp[n][1]));			
	}

	return 0;
}

F. Decreasing Heights

思路：
性質1：當a[1,1]的值確定了，其他的所有的值都將確定

可以肯定的是n*m個數最後肯定存在一個數和之前一樣，沒有改變，那可以枚舉這n*m個數，時間複雜度$O(n^2*n^2)$

代碼：

#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
typedef vector<int> vi;
typedef queue<int> qi;
typedef long long ll;
typedef pair<int, int> pii;
typedef double db;
const int N = 105;
const ll INF = (1LL << 58);
ll dp[N][N], a[N][N];
set<ll> st;

int main()
{
	int t; cin >> t;
	while (t --)
	{
		int n, m; cin >> n >> m;
		st.clear();
		for (int i = 1; i <= n; ++ i)
			for	(int j = 1; j <= m; ++ j)
			{
				cin >> a[i][j];
				st.insert(a[i][j] - (i + j - 2));
			} 
		ll ans = INF;
		for (auto &x : st)
		{
			if(x > a[1][1] || x + (n + m - 2) > a[n][m]) break;
			for (int i = 1; i <= n; ++ i)
				for (int j = 1; j <= m; ++ j)
				{
					ll cost = a[i][j] - (x + (i + j - 2));
					ll val = cost >= 0 ? cost : INF;
					if(cost < 0)
					{
						dp[i][j] = INF;
						continue;
					}
					if(i == 1 && j == 1) dp[i][j] = val;
					else if(i == 1) dp[i][j] = val + dp[i][j - 1];
					else if(j == 1) dp[i][j] = val + dp[i - 1][j];
					else dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + val;
				}
			ans = min(ans, dp[n][m]);
		}
		cout << ans << endl;
	}
	return 0;
}