Codeforces Round #771 (Div. 2) A~E

Posted 2023-02-05 kaka0010

目录

• A. Reverse
• B. Odd Swap Sort
• C. Inversion Graph
• D. Big Brush
• E. Colorful Operations

A. Reverse

给一个$[1..n]$的序列，可以翻转一次，求字典序最小

只需要找到第一个与下标不同的数字，然后把这个数翻转到这个位置即可

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int N = 4e5 + 10;
const int mod = 998244353;

#define fi first
#define se second
#define lowbit(x) x & (-x)

int a[N];
int main() 
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	
	int _; cin >> _; while (_--) 
		int n; cin >> n;
		for (int i = 1; i <= n; i++) cin >> a[i];
		int flag = 0;
		int c = 0;
		for (int i = 1; i <= n; i++) 
			if (i != a[i]) 
				c = i;
				flag = 1;
				break;
			
		
		if (!flag) 
			for (int i = 1; i <= n; i++) cout << a[i] << ' ';
			cout << endl;
			continue;
		
		int d = 0;
		for (int i = 1; i <= n; i++) 
			if (a[i] == c) 
				d = i;
				break;
			
		
		reverse(a+c, a+d+1);
		for (int i = 1; i <= n; i++) cout << a[i] << ' ';
		cout << endl;
	
	return 0;

 

B. Odd Swap Sort

因为只能交换相邻数组，因此需要保证偶数和奇数分别有序

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int N = 4e5 + 10;
const int mod = 998244353;

#define fi first
#define se second
#define lowbit(x) x & (-x)

int a[N], fa[N];

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

	int _; cin >> _; while (_--) 
		int n; cin >> n;
		for (int i = 1; i <= n; i++) cin >> a[i];
		int flag = 0;
		vector<int> c, b;
		for (int i = 1; i <= n; i++) 
			if (a[i] % 2 == 1) c.push_back(a[i]);
		
		for (int i = 1; i <= n; i++) 
			if (a[i] % 2 == 0) b.push_back(a[i]);
		
		flag = 0;
		for (int i = 1; i < b.size(); i++) 
			if (b[i] < b[i-1]) flag = 1;
		
		for (int i = 1; i < c.size(); i++) 
			if (c[i] < c[i-1]) flag = 1;
		
		if (flag) cout << "No" << endl;
		else cout << "Yes" << endl;
	
	
	return 0;

C. Inversion Graph

题意中比较重要的信息，一个是[1…n]的序列，还有一个是逆序对建边

举个例子，3 1 5 2 4
第一个数是3，因此前3个数肯定可以组成一个连通块，到第三个数被更新成5，因此前5个数肯定是一个连通块。。。
还是比较好理解的

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int N = 4e5 + 10;
const int mod = 998244353;

#define fi first
#define se second
#define lowbit(x) x & (-x)

int a[N];
int main() 
	ios::sync_with_stdio(false);
	cin.tie(nullptr);

	int _; cin >> _; while (_--) 
		int n; cin >> n;
		for (int i = 1; i <= n; i++) cin >> a[i];
		int mx = a[1], ans = 1;
		for (int i = 2; i <= n; i++) 
			if (i > mx) 
				ans++;
			 
			mx = max(mx, a[i]);
		
		cout << ans << endl;
	
	 
	return 0;

D. Big Brush

比较暴力的一题，考虑bfs，倒着考虑，把可以涂色的都入队，入队后将当前颜色置位0，为万能色，然后一直暴力找周围的8个点

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int N = 4e5 + 10;
const int mod = 998244353;

#define fi first
#define se second
#define lowbit(x) x & (-x)
#define endl '\\n'

int a[1005][1005], vis[1005][1005];
struct node 
	int x, y, k;
;
int n, m; 
int check(int x, int y) 
	int col = 0;
	int flag = 0;
	for (int i = 0; i < 2; i++) 
		for (int j = 0; j < 2; j++) 
			int dx = x+i;
			int dy = y+j;
			if (dx > n || dy > m) return 0;
			if (a[dx][dy] != col && a[dx][dy] && col) flag = 1;
			if (a[dx][dy]) col = a[dx][dy];  
		
	
	if (flag) return 0;
	else 
		for (int i = 0; i < 2; i++) for (int j = 0; j < 2; j++) a[x+i][y+j]=0;
		return col;
	

int main() 
	ios::sync_with_stdio(false);
	cin.tie(nullptr), cout.tie(nullptr);

	cin >> n >> m;
	queue<node> q;
	for (int i = 1; i <= n; i++) 
		for (int j = 1; j <= m; j++) 
			cin >> a[i][j];
		
	
	for (int i = 1; i <= n; i++) 
		for (int j = 1; j <= m; j++) 
			int cur = check(i, j);
			if (cur) q.push(i,j,cur);
		
	
	vector<node> ans;
	while (q.size()) 
		node now = q.front();
		q.pop();
		if (vis[now.x][now.y]) continue;
		vis[now.x][now.y] = 1;
		// cout << now.x << ' ' << now.y << endl;
		ans.push_back(now);
		for (int l = -1; l <= 1; l++) 
			for (int r = -1; r <= 1; r++) 
				int dx = now.x + l;
				int dy = now.y + r;
				if (dx + 1 <= n && dx >= 1 && dy + 1 <= m && dy >= 1 && !vis[dx][dy]) 
					int cur = check(dx, dy);
					if (cur) q.push(dx,dy,cur);
				
			
		
	
	int flag = 0;
	for (int i = 1; i <= n; i++) 
		for (int j = 1; j <= m; j++) 
			if (a[i][j]) flag = 1;
		
	
	if (flag) 
		cout << "-1" << endl;
		return 0;
	
	cout << ans.size() << endl;
	reverse(ans.begin(), ans.end());
	for (auto it : ans) 
		cout << it.x << ' ' << it.y << ' ' << it.k << endl;
	
	return 0;

E. Colorful Operations

三个操作：1.区间覆盖颜色 2.对某种颜色加权值 3.输出单点权值

对于某种颜色的权值累加和可以开个数组单独处理，然后再修改颜色的时候对权值进行结算，先加上原先颜色的权值，再减去当前颜色的权值

至于怎么处理区间内的颜色，可以用珂朵莉树（其实就是用set模拟暴力）

珂朵莉树详解

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
const int N = 1e6 + 10;
const int mod = 1e9+7;

#define fi first
#define se second
#define lowbit(x) x & (-x)
#define endl '\\n'
#define auto set<node> ::iterator

template <typename T>
struct BIT 
	const int n;
	vector<T> a;
	BIT(int n) :n(n), a(n)
	void add(int x, T v) 
		for (int i = x; i <= n; i += i&-i) 
			a[i] += v;
		
	
	void rangeAdd(int x, int y, T v) 
		add(x, v);
		add(y+1, -v);
	
	T sum(int x) 
		T ans = 0;
		for (int i = x; i > 0; i -= i & -i) 
			ans += a[i];
		
		return ans;
	
	T rangeSum(int l, int r) 
		return sum(r) - sum(l-1);
	
;
struct node 
    int l, r;
    mutable ll val;
    node(int L, int R = -1, ll V = 0):l(L), r(R), val(V) 
    bool operator < (const node &rhs) const 
        return l < rhs.l;
    
;

set<node> odt;
auto split(int pos) 
    auto it = odt.lower_bound(node(pos));
    if (it != odt.end() && it->l == pos) return it;
    it--;
    int L = it->l, R = it->r;
    ll val = it->val;
    odt.erase(it);
    odt.insert(node(L, pos-1, val));
    return odt.insert(node(pos, R, val)).first;


vector<ll> add(N);
BIT<ll> bit(N);

void assignCodeforces Round #771 (Div. 2)（ABCDE）

 Codeforces Round #771 (Div. 2)（ABCDE）
 Codeforces Round #771 (Div. 2) A~E
 Codeforces Round #771 (Div. 2) A~E
 Codeforces Round #771 (Div. 2) A~E
 Codeforces Round #436 E. Fire（背包dp+输出路径）