坐牢一个半小时...D、E待补(太菜了，做的题还是太少了）

A - Constructive 问题集 

        

 思路：手画一下发现：n个城市最多能重建n * n 的城市，所以n * m 需要重建max(n , m)个城市。

// Problem: A. Constructive Problems
// Contest: Codeforces - Codeforces Round 915 (Div. 2)
// URL: https://codeforces.com/contest/1905/problem/0
// Memory Limit: 256 MB
// Time Limit: 1000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define pb push_back
#define x first
#define y second 
#define endl '\n'
const LL maxn = 4e05+7;
const LL N = 5e05+10;
const LL mod = 1e09+7;
const int inf = 0x3f3f3f3f;
const LL llinf = 5e18;
typedef pair<int,int>pl;
priority_queue<LL , vector<LL>, greater<LL> >mi;//小根堆
priority_queue<LL> ma;//大根堆
LL gcd(LL a, LL b){
	return b > 0 ? gcd(b , a % b) : a;
}

LL lcm(LL a , LL b){
	return a / gcd(a , b) * b;
}
int n , m;
vector<int>a(N , 0);
void init(int n){
	for(int i = 0 ; i <= n ; i ++){
		a[i] = 0;
	}
}
void solve() 
{
	cin >> n >> m;
	cout << max(n , m) << endl;	
}            
int main() 
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cout.precision(10);
    int t=1;
	cin>>t;
    while(t--)
    {
    	solve();
    }
    return 0;
}

B - Begginer's Zelda 

        题意：给定一棵树，能够进行如下操作：选择两个点u , v 。将u到v的简单路径上的所有点缩成一个点，并将与简单路径上相连的边连到那个点上面。

        

思路：最佳策略必然是选择两个叶子结点，即度为1的点。这样每次操作就能够使得结点数减少2。

       因此最小次数为（叶子结点个数 + 1） /  2。

// Problem: B. Begginer's Zelda
// Contest: Codeforces - Codeforces Round 915 (Div. 2)
// URL: https://codeforces.com/contest/1905/problem/B
// Memory Limit: 256 MB
// Time Limit: 1000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define pb push_back
#define x first
#define y second 
#define endl '\n'
const LL maxn = 4e05+7;
const LL N = 5e05+10;
const LL mod = 1e09+7;
const int inf = 0x3f3f3f3f;
const LL llinf = 5e18;
typedef pair<int,int>pl;
priority_queue<LL , vector<LL>, greater<LL> >mi;//小根堆
priority_queue<LL> ma;//大根堆
LL gcd(LL a, LL b){
	return b > 0 ? gcd(b , a % b) : a;
}

LL lcm(LL a , LL b){
	return a / gcd(a , b) * b;
}
int n , m;
vector<int>a(N , 0);
void init(int n){
	for(int i = 0 ; i <= n ; i ++){
		a[i] = 0;
	}
}
void solve() 
{
	cin >> n;
	vector<int>deg(n + 5 , 0);
	for(int i = 1 ; i < n ; i ++){
		int u , v;
		cin >> u >> v;
		deg[u]++;
		deg[v]++;
	}		
	int cnt = 0;
	for(int i = 1 ; i <= n ; i ++){
		if(deg[i] == 1){
			cnt ++;
		}
	}
	cout << (cnt + 1) / 2 << endl;
}            
int main() 
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cout.precision(10);
    int t=1;
	cin>>t;
    while(t--)
    {
    	solve();
    }
    return 0;
}

C - Largest Subsequence 

        题意：给定一个字符串，每次操作将选中字符串中最大子序列，并且将他们相对位置循环右移。求最终字符串有序的最小操作数，或输出-1。

        

题意：考虑如何选最大子序列：按照字母大小从大到小选，当前字母选完之后才选下一个字母，而下一个字母从前一个字母的右端开始选。

        然后发现：每一轮操作选择的子序列都是前一个操作的子集（因为最后一个到了第一位，不再选择了，其余都照常选）。因此最终的操作数即为使得第一轮选择的子序列变为有序的操作数。

然后再考虑最终操作完了能否使得整个字符串有序即可。

        

// Problem: C. Largest Subsequence
// Contest: Codeforces - Codeforces Round 915 (Div. 2)
// URL: https://codeforces.com/contest/1905/problem/C
// Memory Limit: 256 MB
// Time Limit: 1000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define pb push_back
#define x first
#define y second 
#define endl '\n'
const LL maxn = 4e05+7;
const LL N = 5e05+10;
const LL mod = 1e09+7;
const int inf = 0x3f3f3f3f;
const LL llinf = 5e18;
typedef pair<int,int>pl;
priority_queue<LL , vector<LL>, greater<LL> >mi;//小根堆
priority_queue<LL> ma;//大根堆
LL gcd(LL a, LL b){
	return b > 0 ? gcd(b , a % b) : a;
}

LL lcm(LL a , LL b){
	return a / gcd(a , b) * b;
}
int n , m;
vector<int>a(N , 0);
void init(int n){
	for(int i = 0 ; i <= n ; i ++){
		a[i] = 0;
	}
}
// z n m i g e c
void solve() 
{
	set<int>pos[26];
	cin >> n;
	string s;
	cin >> s;
	vector<int>change;
	vector<int>c(n , 0);
	vector<int>vis(n , 1);
	for(int i = 0 ; i < n ; i ++){
		a[i] = s[i] - 'a';
		pos[a[i]].insert(i);
	}
	int st = -1;
	for(int i = 25 ; i >= 0 ;i --){
		for(auto it : pos[i]){
			if(it > st){
				change.pb(it);
				vis[it] = 0;
				st = it;
			}
		}
	}
	int num = 0;
	int len = change.size();
	for(int i = 0 ; i < len ; i ++){
		if(a[change[0]] == a[change[i]])
			num++;
	}
	int op = change.size() - num;
	int l = change.size() - 1;
	for(int i = 0 ; i < n ; i ++){
		if(vis[i]){
			c[i] = a[i];
		}
		else{
			c[i] = a[change[l]];
			l--;
		}
	}
	for(int i = 1 ; i < n ; i ++){
		if(c[i] < c[i - 1]){
			cout << -1 << endl;
			return;
		}
	}
	cout << op << endl;
}            
int main() 
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cout.precision(10);
    int t=1;
	cin>>t;
    while(t--)
    {
    	solve();
    }
    return 0;
}