The 2022 ICPC Asia Regionals Online Contest (I) C L A

The 2022 ICPC Asia Regionals Online Contest (I)

C

统计度的大小，算贡献，特判 \(n = 1\)

#include<bits/stdc++.h>

using namespace std;

const int N = 1e6 + 10;

typedef long long ll;

int n, d[N];
vector<int> e[N];
ll res = 0;
void dfs(int u, int from)
{
	int cnt = 0;
	for(auto v : e[u])
	{
		if(v == from)	continue;
		cnt++;
		dfs(v, u);
	}
	if(u != 1)
		res = res + max(cnt - 1, 0);
}

void solve()
{
	res = 0;
	cin>>n;
	if(n == 1)
	{
		cout<<1<<'\n';
		return;
	}
	for(int i = 1; i <= n; i++)
	{
		d[i] = 0;
		e[i].clear();
	}
	for(int i = 1; i <= n - 1; i++)
	{
		int u, v;	cin>>u>>v;
		e[u].push_back(v);
		e[v].push_back(u);
		d[u]++, d[v]++;
	}
	dfs(1, 0);
	res += 2;
	if(d[1] >= 2)
		res += (d[1] - 2);
	cout<<res<<'\n';
}


int main()
{
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	cout.tie(nullptr);
	int tc;	cin>>tc;
	while(tc--)
		solve();

}

L

动态规划

预处理所有不合法情况

#include<bits/stdc++.h>

using namespace std;

const int N = 5e5 + 10;

typedef long long ll;

int n, m, res;
int a[N], b[N], f[N][27], g[N][27];
string s, t;    
bool st[N][27], vis[27];
void solve()
{
    cin>>s>>t;
    n = s.size(), m = t.size();
    s = "?" + s, t = "?" + t;
    for(int i = 1; i <= n; i++)
        a[i] = s[i] - 'a' + 1;
    for(int i = m; i >= 1; i--)
    {
        b[i] = t[i] - 'a' + 1;
        for(int j = 1; j <= 26; j++)
            if(vis[j]) st[b[i]][j] = true;
        vis[b[i]] = true;
    }
    for(int i = 1; i <= n; i++)
    {
        for(int j = 1; j <= 26; j++)
            f[i][j] = f[i - 1][j] + (vis[a[i]] == false), res = max(f[i][j], res);
        if(vis[a[i]] == false)  continue;   
        f[i][a[i]] = max(1, f[i][a[i]]);
        res = max(f[i][a[i]], res);
        for(int j = 1; j <= 26; j++)
            if(st[j][a[i]] == false)
                f[i][a[i]] = max(f[i - 1][j] + 1, f[i][a[i]]),
                res = max(f[i][a[i]], res);
    }
    cout<<res<<'\n';
}


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

}

A

观察得：

1. 一团大小为\(k\) 的 \(1\) 可以删掉 \(\dfrac{k}{2} + k \% 2\)

2. 对于 \(1,1,0,\dots,0,1,1\), 首尾的 \(1\) 会相互影响，但对于记录一下其前后缀就可以得出首尾的 一团 \(1\) 的大小

3. 那么预处理出删的次数即可

   const int N = 1e6 + 10;
   
   int n, q, pre[N], suf[N], f[N], cnt;
   string s;
   void solve()
   {       
       cin>>n>>q;
       cin>>s; s = "?" + s;
       for(int i = 1; i <= n; i++)
       {
           if(s[i] == '1') cnt++;
           else cnt = 0;
           pre[i] = cnt;
           f[i] = f[i - cnt - 1] + (cnt / 2 + cnt % 2);
       }
       cnt = 0;
       for(int i = n; i >= 1; i--)
       {
           if(s[i] == '1') cnt++;
           else cnt = 0;
           suf[i] = cnt;
       }
       for(int i = 1; i <= q; i++)
       {
           int l, r;   cin>>l>>r;
           int tl = l + suf[l], tr = r - pre[r];
           if(suf[l] == 0 && pre[r] != 0)
               tl++;
           if(pre[r] == 0 && suf[l] != 0)
               tr--;
           int t = suf[l] + pre[r];
           t = t / 2 + t % 2;
           int add = 0;
           if(tl <= tr)
               add = f[tr] - f[tl - 1];
           int res = (r - l + 1) / 3 - t - add;
           res = max(0, res);
           cout<<res<<'\n';
       }
       return;
   }