2022 NOIP 补题
P8865 [NOIP2022] 种花
注意到 \(F\) 的构造仅仅在 \(C\) 上多了一个横
那么我们统计 \(F\) 的时候统计 \(C\) 即可
考虑预处理对于每一个点向右扩展的合法横边数量 用前缀和实现即可
考虑对 \(C\) 统计 对于 \((a,b)\) 点作为左下角的统计 记录 \(cntl\) 为上面连续(需要隔一行)可达的合法的横边数量
例如我们对下面的东西统计时
00011
01111
00001
00001
在第 \(4\) 行的时候 \(cntl\) 为 \(2\) (因为上一行的边不可达 否则为 \(5\) )
同时 \(cntc\) 为之前可达的所有 \(C\) 图形个数 这样我们累加 \(F\) 即可
所以做完了()
#include<bits/stdc++.h>
using namespace std;
#define mid ((l+r)>>1)
#define inl inline
#define eb emplace_back
#define endl '\n'
#define pii pair<int,int>
#define mkp make_pair
#define print(x) cerr<<#x<<'='<<x<<endl
#define getchar() cin.get()
const int N = 1e3 + 5;
const int inf = 0x3f3f3f3f;
const int mod = 998244353;
int read()
{
int x = 0 , f = 1;
char ch = getchar();
while ( !isdigit(ch) ) { if ( ch == '-' ) f = -1; ch = getchar(); }
while ( isdigit(ch) ) { x = ( x << 1 ) + ( x << 3 ) + ( ch ^ 48 ); ch = getchar(); }
return x * f;
}
int n , m , C , F , f[N][N] , ansc , ansf;
char a[N][N];
signed main ()
{
ios::sync_with_stdio(false);
cin.tie(0) , cout.tie(0);
int T = read() , id = read();
while ( T -- )
{
memset ( f , 0 , sizeof f ) , ansc = ansf = 0;
n = read() , m = read() , C = read() , F = read();
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 = m - 1 ; j ; j -- )
{
if ( a[i][j] == '1' ) f[i][j] = -1;
else if ( a[i][j+1] == '0' ) f[i][j] = f[i][j+1] + 1;
}
for ( int j = 1 ; j < m ; j ++ )//最后一列显然不用枚举
{
int cntl = 0 , cntc = 0;
for ( int i = 1 ; i <= n ; i ++ )
{
if ( f[i][j] == -1 ) { cntl = cntc = 0; continue; }
( ansc += f[i][j] * cntl % mod ) %= mod;
( ansf += cntc ) %= mod;
( cntc += f[i][j] * cntl % mod ) %= mod;
( cntl += max ( 0 , f[i-1][j] ) ) %= mod;
}
}
cout << C * ansc % mod << ' ' << F * ansf % mod << endl;
}
return 0;
}
P8866 [NOIP2022] 喵了个喵
对于 \(k=2n-2\)，显然我们最多是将前 \(n-1\) 个栈全部塞满不同的数，那么我们如果再加进来一个数，是一定可以消除的。
那么我们留栈 \(n\) 为辅助栈。
对于 \(1\) 操作，显然是扫一遍栈来找到一个顶端相同的栈塞进去。
对于 \(2\) 操作，我们将当前数加到栈 \(n\) 中，然后扫一遍找到相同的即可。
否则直接塞进第一个大小 \(< 2\) 的栈即可。
#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define inl inline
#define eb emplace_back
#define pb pop_back
#define getchar() cin.get()
#define pii pair<int,int>
#define mkp make_pair
#define fi first
#define se second
const int M = 2e6 + 5;
const int N = 300 + 5;
int read()
{
int f = 1 , x = 0;
char ch = getchar();
while ( !isdigit(ch) ) { if ( ch == '-' ) f = -1; ch = getchar(); }
while ( isdigit(ch) ) { x = ( x << 1 ) + ( x << 3 ) + ( ch ^ 48 ); ch = getchar(); }
return x * f;
}
int n , m , k , T;
namespace sub1
{
int a[M];
struct node { int op , x , y; };
vector<node> ans;
deque<int> sta[N];
void main()
{
while ( T -- )
{
n = read() , m = read() , k = read();
for ( int i = 1 ; i <= n ; i ++ ) sta[i].clear();
ans.clear();
for ( int i = 1 ; i <= m ; i ++ )
{
a[i] = read();
int flag = 0;
for ( int j = 1 ; j < n ; j ++ )
if ( sta[j].size() )
{
if ( sta[j].back() == a[i] ) { ans.eb ( (node) { 1 , j , 0 } ) , sta[j].pb() , flag = 1; break; }
else if ( sta[j].front() == a[i] ) { ans.eb ( (node) { 1 , n , 0 } ) , ans.eb ( (node) { 2 , j , n } ) , sta[j].pop_front() , flag = 1; break; }
}
if ( !flag )
{
for ( int j = 1 ; j < n ; j ++ ) if ( sta[j].size() < 2 ) { sta[j].push_back(a[i]); ans.eb ( (node) { 1 , j , 0 } ); break; }
}
}
cout << ans.size() << endl;
for ( auto p : ans )
{
if ( p.op == 1 ) cout << p.op << ' ' << p.x << endl;
else cout << p.op << ' ' << p.x << ' ' << p.y << endl;
}
}
}
}
namespace sub2
{
int a[M] , flag;
struct node { int op , x , y; };
vector<node> ans , temp;
deque<int> sta[N];
int check()
{
// cout << "OK" << sta[1].size() << ' ' << sta[2].size() << endl;
// cout << "ans" << endl;
// for ( auto p : ans )
// {
// if ( p.op == 1 ) cout << p.op << ' ' << p.x << endl;
// else cout << p.op << ' ' << p.x << ' ' << p.y << endl;
// }
// cout << "ans" << endl;
for ( int i = 1 ; i <= n ; i ++ ) if ( sta[i].size() != 0 ) return 0;
return 1;
}
void dfs ( int stp , int lim , int now )
{
// cout << stp << ' ' << lim << ' ' << now << endl;
if ( stp > lim + 1 || now > m + 1 ) return;
if ( stp == lim + 1 )
{
if ( now == m + 1 && check() ) temp = ans , flag = 1;
return;
}
for ( int j = 1 ; j <= n ; j ++ )
if ( sta[j].size() && sta[j].back() == a[now] )
{
ans.eb ( (node) { 1 , j , 0 } ) , sta[j].pb();
dfs ( stp + 1 , lim , now + 1 );
ans.pb() , sta[j].eb(a[now]);
if ( flag ) return;
}
else
{
ans.eb ( (node) { 1 , j , 0 } ) , sta[j].eb(a[now]) , dfs ( stp + 1 , lim , now + 1 ) , sta[j].pb() , ans.pb();
if ( flag ) return;
}
for ( int i = 1 ; i <= n ; i ++ )
for ( int j = i + 1 ; j <= n ; j ++ )
if ( sta[i].size() && sta[j].size() && sta[i].front() == sta[j].front() )
{
int ti = sta[i].front() , tj = sta[j].front();
ans.eb ( (node) { 2 , i , j } ) , sta[i].pop_front() , sta[j].pop_front();
dfs ( stp + 1 , lim , now );
ans.pb() , sta[i].push_front(ti) , sta[j].push_front(tj);
if ( flag ) return;
}
}
void main()
{
while ( T -- )
{
n = read() , m = read() , k = read();
for ( int i = 1 ; i <= n ; i ++ ) sta[i].clear();
temp.clear() , ans.clear();
for ( int i = 1 ; i <= m ; i ++ ) a[i] = read();
for ( int i = m ; i <= 2 * m ; i ++ )
{
flag = 0 , dfs ( 1 , i , 1 );
if ( flag ) break;
}
cout << temp.size() << endl;
for ( auto p : temp )
{
if ( p.op == 1 ) cout << p.op << ' ' << p.x << endl;
else cout << p.op << ' ' << p.x << ' ' << p.y << endl;
}
}
}
}
signed main ()
{
ios::sync_with_stdio(false);
cin.tie(0) , cout.tie(0);
T = read();
if ( T % 10 == 1 ) sub1::main();
else sub2::main();
return 0;
}
/*
2
2 4 2
1 2 1 2
2 4 2
1 2 1 2
*/
P8867 [NOIP2022] 建造军营
\(O(2^nm(n+m))\) 的搜索。
#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define inl inline
#define eb emplace_back
#define pb pop_back
#define getchar() cin.get()
#define pii pair<int,int>
#define mkp make_pair
#define fi first
#define se second
const int N = 2e4 + 5;
const int mod = 1e9 + 7;
int read()
{
int f = 1 , x = 0;
char ch = getchar();
while ( !isdigit(ch) ) { if ( ch == '-' ) f = -1; ch = getchar(); }
while ( isdigit(ch) ) { x = ( x << 1 ) + ( x << 3 ) + ( ch ^ 48 ); ch = getchar(); }
return x * f;
}
int n , m , cnt , ans;
struct edge { int u , v; } e[N];
inl void add ( int u , int v ) { e[++cnt] = { u , v }; }
vector<int> temp;
struct Dsu
{
int fa[N];
void init() { for ( int i = 1 ; i <= n ; i ++ ) fa[i] = i; }
int find ( int x ) { return fa[x] == x ? x : fa[x] = find(fa[x]); }
void merge ( int u , int v )
{
int fu = find(u) , fv = find(v);
if ( fu != fv ) fa[fu] = fv;
}
int pd ()
{
int fir = 0;
for ( int i = 1 ; i <= n ; i ++ ) if ( temp[i] == 1 ) { fir = i; break; }
for ( int i = 1 ; i <= n ; i ++ ) if ( temp[i] == 1 && find(i) != find(fir) ) return 0;
return 1;
}
}D;
void solve ()
{
int fir = 0;
for ( int i = 1 ; i <= n ; i ++ ) if ( temp[i] == 1 ) { fir = i; break; }
if ( !fir ) return;
int res = 0;
for ( int i = 1 ; i <= m ; i ++ )
{
D.init();
for ( int j = 1 ; j <= m ; j ++ ) if ( j != i ) D.merge ( e[j].u , e[j].v );
res += D.pd();
}
( ans += ( 1 << res ) ) %= mod;
}
void dfs ( int stp )
{
if ( stp == n + 1 ) return solve();
temp.eb(0) , dfs ( stp + 1 ) , temp.pb();
temp.eb(1) , dfs ( stp + 1 ) , temp.pb();
}
signed main ()
{
// ios::sync_with_stdio(false);
// cin.tie(0) , cout.tie(0);
n = read() , m = read();
for ( int i = 1 , u , v ; i <= m ; i ++ ) u = read() , v = read() , add ( u , v );
temp.eb(0) , dfs ( 1 );
cout << ans << endl;
return 0;
}
P8868 [NOIP2022] 比赛
\(8pts\ O(n^4q)\) 太显然了。
考虑优化这个东西。
我们将询问挂到 \(r\) 端点上。
那么枚举 \(r\)，设 \(f_i\) 表示的是 \([i,r]\) 之间的答案总和，\(x_i,y_i\) 表示的是 \([i,r]\) 之间的 \(a,b\) 数组最大值。
那么对于新加入的 \(r\)，\([1,r]\) 之间的所有 \(x,y,f\) 都可以 \(O(1)\) 维护，处理后统计询问即可，询问的答案为 \(\sum_{j=i}^r f_i\)。
这样的时间复杂度为 \(O(n^2+nq)\)。
#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define inl inline
#define eb emplace_back
#define pb pop_back
#define getchar() cin.get()
#define pii pair<int,int>
#define mkp make_pair
#define fi first
#define se second
#define int unsigned long long
const int N = 3e5 + 5;
int read()
{
int f = 1 , x = 0;
char ch = getchar();
while ( !isdigit(ch) ) { if ( ch == '-' ) f = -1; ch = getchar(); }
while ( isdigit(ch) ) { x = ( x << 1 ) + ( x << 3 ) + ( ch ^ 48 ); ch = getchar(); }
return x * f;
}
int n , a[N] , b[N] , q , x[N] , y[N] , ans[N] , f[N];
vector<pii> que[N];
signed main ()
{
ios::sync_with_stdio(false);
cin.tie(0) , cout.tie(0);
int T = read();
n = read();
for ( int i = 1 ; i <= n ; i ++ ) a[i] = read();
for ( int i = 1 ; i <= n ; i ++ ) b[i] = read();
q = read();
for ( int i = 1 , l , r ; i <= q ; i ++ ) l = read() , r = read() , que[r].eb(l,i);
for ( int r = 1 ; r <= n ; r ++ )
{
for ( int i = 1 ; i <= r ; i ++ ) x[i] = max ( x[i] , a[r] ) , y[i] = max ( y[i] , b[r] ) , f[i] += x[i] * y[i];
for ( auto p : que[r] )
{
int l = p.fi , id = p.se;
for ( int i = l ; i <= r ; i ++ ) ans[id] += f[i];
}
}
for ( int i = 1 ; i <= q ; i ++ ) cout << ans[i] << endl;
return 0;
}
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