题解 | Birthday Reminders-2019牛客暑期多校训练营第九场F题

题目来源于牛客竞赛：https://ac.nowcoder.com/acm/contest/discuss
题目描述：

输入描述：

输出描述：

示例1：

题解：

代码：

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>using namespace std;
using namespace __gnu_pbds;#define fi first
#define se second
#define mp make_pair
#define pb push_backtypedef long long ll;
typedef pair<ll,ll> ii;
typedef vector<int> vi;
typedef unsigned long long ull;
typedef long double ld;
typedef tree<ii, null_type, less<ii>, rb_tree_tag, tree_order_statistics_node_update> pbds;vector<int> fact;
vector<int> ifact;
vector<int> inv;
vector<int> pow2;
const int MOD = (1e9 + 7);
int add(int a, int b)
{a+=b;while(a>=MOD) a-=MOD;return a;
}
int mult(int a, int b)
{return (a*1LL*b)%MOD;
}
int modpow(int a, int b)
{int r=1;while(b){if(b&1) r=mult(r,a);a=mult(a,a);b>>=1;}return r;
}
int choose(int a, int b)
{if(a<b) return 0;if(b==0) return 1;if(a==b) return 1;return mult(fact[a],mult(ifact[b],ifact[a-b]));
}
int inverse(int a)
{return modpow(a,MOD-2);
}
void init(int _n)
{fact.clear(); ifact.clear(); inv.clear(); pow2.clear();fact.resize(_n+1);ifact.resize(_n+1);inv.resize(_n+1);pow2.resize(_n+1);pow2[0]=1;ifact[0]=1;fact[0]=1;for(int i=1;i<=_n;i++){pow2[i]=add(pow2[i-1],pow2[i-1]);fact[i]=mult(fact[i-1],i);//ifact[i]=mult(ifact[i-1],inv[i]);}ifact[_n] = inverse(fact[_n]);for(int i=_n-1;i>=1;i--){ifact[i] = mult(ifact[i + 1], i + 1);}for(int i=1;i<=_n;i++){inv[i] = mult(fact[i-1],ifact[i]);}
}int cnt[13];
vi v;
vi perm;void dfs(int u)
{if(cnt[u]>=3) return ;cnt[u]++;v[perm[u]]=min(v[perm[u]],v[u]+1);dfs(perm[u]);
}int solve_naive(vi t)
{int n=t.size();for(int i=0;i<n;i++){if(t[i]==-1) t[i]=int(1e9);}perm.resize(n);set<vi> S;for(int i=0;i<n;i++) perm[i]=i;do{v=t;memset(cnt,0,sizeof(cnt));for(int i=0;i<n;i++){if(cnt[i]==0) dfs(i);}sort(v.begin(),v.end());for(int i=0;i<n;i++){if(v[i]>=int(1e9)) v[i]=-1;assert(v[i]<10000);}//for(int x:v) cerr<<x<<' ';//cerr<<'\n';S.insert(v);}while(next_permutation(perm.begin(),perm.end()));return S.size();
}const int C = 200;
int dp[C+2][C+2][C+2]; //cur node, sum, last, # of badvoid radd(int &a, int b)
{a=add(a,b);
}int solve_fast(vi t)
{memset(dp,0,sizeof(dp));vi cnt(C,0);vi pref(C,0);int n=t.size();for(int i=0;i<t.size();i++){if(t[i]==-1) t[i]=C-1;assert(t[i]>=0&&t[i]<C);cnt[t[i]]++;}for(int i=1;i<C;i++) pref[i]=cnt[i]+pref[i-1];dp[0][0][0]=1;for(int i=0;i<C;i++){for(int j=0;j<=n;j++){for(int k=0;k<=n;k++){if(dp[i][j][k]==0) continue;int v=dp[i][j][k];for(int c=0;j+c<=n;c++){int realc = cnt[i+1];int newbad=j-pref[i];if(c>=realc){//need to borrow from front//we need c-realc more to hook with the frontint needed=c-realc;if(k<needed) continue;//exist >= needed to hook with//# of bad person increases by needednewbad+=needed;}else{//need to finally "use up the needed ones"int extra = realc-c;if(extra>newbad) continue;//extra<=lnewbad-=extra;}radd(dp[i+1][j+c][c],v);}}}}int ans=0;for(int k=0;k<=n;k++){radd(ans,dp[C-1][n][k]);}return ans;
}void read_solve()
{int n; cin>>n;vi t(n);for(int i=0;i<n;i++) cin>>t[i];//int naive = solve_naive(t);int fast = solve_fast(t);cout<<fast<<'\n';
}void gen_solve()
{int n = rand()%8+1;vi t(n);int p=rand()%100+1;int C=rand()%6+1;for(int i=0;i<n;i++){t[i]=rand()%C+1;if(rand()%100<=p) t[i]=-1;}int naive = solve_naive(t);cerr<<"NAIVE : "<<naive<<'\n';int fast = solve_fast(t);cerr<<"FAST : "<<fast<<'\n';if(naive!=fast){freopen("birthday-wishes.in","w",stdout);cout<<n<<'\n';for(int i=0;i<n;i++){cout<<t[i]<<' ';}cout<<'\n';exit(0);}
}int main()
{ios_base::sync_with_stdio(0); cin.tie(0);init(123333);read_solve();
}

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