noip20

T1

首先，这个插球，就是森林中连点，考虑dp，我们设 \(dp_{i,j}\) 表示有i个点的森林，有j个点在第一颗树中的概率，转移时考虑第i个点是否在第一颗子树里，转移方程：

\[dp_{i,j}=dp_{i-1,j-1}\times (j-1)\times inv_{i}+dp_{i-1,j}\times (i-j)\times inv_{i}
\]

设 \(f_{i,j}\) 表示有i个的树，深度不超过j的概率，\(g_{i,j}\) 表示有i个点的森林，深度不超过j的概率，转移方程为：

\[f_{i,j}=g_{i-1,j-1}
\]

\[g_{i,j}=\sum_{k=1}^{i} f_{k,j}\times g_{i-k,j}\times dp_{i,k}
\]

比较恶心的是初始化，按照定义来初始化就好，就是容易漏情况，具体看code。

注意取模QAQ

Code

#include<cstdio>
#define N 220
#define re register
#define int long long
namespace OMA
{
   int n,p,inv[N],sum;
   int dp[N][N],f[N][N],g[N][N];
   inline int quickpow(int a,int b)
   {
     int ans = 1;
     while(b)
     {
       if(b&1)
       { ans = ans*a%p; }
       a = a*a%p;
       b >>= 1;
     }
     return ans;
   }
   signed main()
   {
     scanf("%lld%lld",&n,&p);
     dp[1][1] = dp[1][0] = 1;
     inv[0] = 1,inv[n] = quickpow(n,p-2);
     for(re int i=n-1; i>=1; i--)
     { inv[i] = quickpow(i,p-2); }
     for(re int i=2; i<=n; i++)
     {
       for(re int j=1; j<=i; j++)
       { dp[i][j] = ((dp[i-1][j-1]*(j-1)%p*inv[i]%p+dp[i-1][j]*(i-j)%p*inv[i]%p)%p+p)%p; }
     }
     for(re int i=0; i<=n; i++)
     { f[0][i] = g[0][i] = 1; }
     for(re int i=1; i<=n; i++)
     {
       for(re int j=i-1; j<=n; j++)
       { f[i][j] = g[i][j] = 1; }
     }
     for(re int i=2; i<=n; i++)
     {
       for(re int j=0; j<=i-2; j++)
       {
         if(j)
         { f[i][j] = g[i-1][j-1]; }
         for(re int k=1; k<=i; k++)
         { (g[i][j] += f[k][j]*g[i-k][j]%p*dp[i][k]%p) %= p; }
       }
     }
     for(re int i=1; i<=n; i++)
     { (sum += ((f[n][i]-f[n][i-1])*i%p+p)%p) %= p; }
     printf("%lld\n",sum);
     return 0;
   }
}
signed main()
{ return OMA::main(); }

T2

考试的时候想骗 \(c=0\) 的点，结果挂了，

90 1 19

16 79 0

傻逼数据，1不连边

记搜有72pts..

TLE 72pts

#include<bitset>
#include<cstdio>
#define re register
const int N = 111;
const int M = 8500;
namespace OMA
{
   int n,m,d,ans;
   struct graph
   {
     int next;
     int to;
     int w;
   }edge[M<<1];
   int staa[1<<21];
   std::bitset<1<<20>vis[N][N];
   int cnt=1,head[N];
   inline int read()
   {
     int s=0,w=1; char ch=getchar();
     while(ch<'0'||ch>'9'){ if(ch=='-')w=-1; ch=getchar(); }
     while(ch>='0'&&ch<='9'){ s=s*10+ch-'0'; ch=getchar(); }
     return s*w;
   }
   inline void add(int u,int v,int w)
   {
     edge[++cnt].next = head[u];
     edge[cnt].to = v;
     edge[cnt].w = w;
     head[u] = cnt;
   }
   inline void dfs(int u,int sta,int dnt)
   {
     if(vis[u][dnt][sta])
     { return ; }
     if(dnt==d)
     { if(!staa[sta]){ ans++,staa[sta] = 1; } return ; }
     vis[u][dnt][sta] = 1;
     for(re int i=head[u]; i; i=edge[i].next)
     {
       int v = edge[i].to,w = edge[i].w;
       int stat = sta|(w<<dnt);
       dfs(v,stat,dnt+1);
     }
   }
   signed main()
   {
     n = read(),m = read(),d = read();
     for(re int i=1,u,v,w; i<=m; i++)
     {
       u = read(),v = read(),w = read();
       add(u,v,w),add(v,u,w);
     }
     dfs(1,0,0);
     printf("%d\n",ans);
     return 0;
   }
}
signed main()
{ return OMA::main(); }

正解：

显然是状压，设 \(dp_{i,j,mask}\) 表示从i出发，到j结束，是否存在一条状态为mask的路径，最后枚举转移答案，但直接转移会T掉题解里说的，考虑如何优化

meet in the middle+bitset

具体实现可见代码。

时间复杂度 \(O(2^{\frac{d}{2}}\times n\times (n+m)+2^{d}\times n)\)

也是题解里说的

Code

#include<bitset>
#include<cstdio>
#include<cstring>
#define re register
const int N = 221;
const int M = 8500;
const int MASK = 1<<10+1;
namespace OMA
{
   int n,m,d[3],ans;
   std::bitset<N>dp[MASK][2],link[N][2];
   inline int read()
   {
     int s=0,w=1; char ch=getchar();
     while(ch<'0'||ch>'9'){ if(ch=='-')w=-1; ch=getchar(); }
     while(ch>='0'&&ch<='9'){ s=s*10+ch-'0'; ch=getchar(); }
     return s*w;
   }
   signed main()
   {
     n = read(),m = read(),d[0] = read();
     for(re int i=1,u,v,w; i<=m; i++)
     {
       u = read(),v = read(),w = read();
       link[u][w][v] = link[v][w][u] = 1;
     }
     d[2] = d[0]-(d[1] = d[0]/2);
     int mask[2] = {1<<d[1],1<<d[2]};
     for(re int u=n; u; u--)
     {
       for(re int i=0; i<MASK; i++)
       { dp[i][0].reset(); }
       dp[1][0][u] = 1;
       for(re int i=0; i<=mask[1]-1; i++)
       {
         for(re int v=1; v<=n; v++)
         {
           if(dp[i][0][v])
           { dp[i<<1][0] |= link[v][0],dp[i<<1|1][0] |= link[v][1]; }
         }
       }
       for(re int i=0; i<=mask[1]-1; i++)
       { dp[i][1][u] = dp[mask[1]|i][0].any(); }
     }
     for(re int i=0; i<=mask[1]-1; i++)
     {
       for(re int j=0; j<=mask[0]-1; j++)
       {
         if((dp[mask[0]|j][0]&dp[i][1]).any())
         { ans++; }
       }
     }
     printf("%d\n",ans);
     return 0;
   }
}
signed main()
{ return OMA::main(); }

T3

暴力30pts+ \(q=0\) 1pts+单调14pts

单调包括递增和递减，注意一下。

45pts

#include<cmath>
#include<cstdio>
#define MAX 100100
#define re register
#define int long long
namespace OMA
{
   int n,q;
   int x[MAX],len[MAX];
   inline int read()
   {
     int s=0,w=1; char ch=getchar();
     while(ch<'0'||ch>'9'){ if(ch=='-')w=-1; ch=getchar(); }
     while(ch>='0'&&ch<='9'){ s=s*10+ch-'0'; ch=getchar(); }
     return s*w;
   }
   inline int abs(int a)
   { return a>=0?a:-a; }
   inline void task1()
   {
     int l,r;
     long long cast;
     for(re int i=1; i<=q; i++)
     {
       cast = l = 0,r = len[i];
       for(re int j=1; j<=n; j++)
       {
         //printf("l=%d r=%d to=%d ",l,r,x[j]);
         if((r>x[j]&&l<x[j])||(r==x[j])||(l==x[j]))
         { continue ; }
         else if(abs(r-x[j])<abs(l-x[j]))
         { cast += abs(r-x[j]),r = x[j],l = r-len[i]; }
         else
         { cast += abs(l-x[j]),l = x[j],r = l+len[i]; }
         //printf("haven casted:%d\n",cast);
       }
       printf("%lld\n",cast);
     }
   }
   inline void task2()
   {
     int cnt[2] = {0,0}; // up down
     for(re int i=2; i<=n; i++)
     {
       if(x[i]>x[i-1])
       { cnt[0]++; }
       else if(x[i]<x[i-1])
       { cnt[1]++; }
     }
     if(cnt[0]==n-1)
     {
       if(x[1]>=0)
       {
         for(re int i=1; i<=q; i++)
         { printf("%lld\n",(len[i]<x[n])?(x[n]-len[i]):0); }
       }
       else
       {
         for(re int i=1; i<=q; i++)
         { printf("%lld\n",abs(x[1])+abs(x[n]-x[1]-len[i])); }
       }
     }
     if(cnt[1]==n-1)
     {
       for(re int i=1; i<=q; i++)
       {
         int cast = 0;
         if(x[1]>len[i])
         { cast = x[1]-len[i]; }
         if(x[n]<cast)
         { cast += cast-x[n]; }
         printf("%lld\n",cast);
       }
     }
   }
   inline void task3()
   { ; }
   inline void task4()
   { ; }
   signed main()
   {
     //freopen("node.in","r",stdin);
     //freopen("my.out","w",stdout);
     n = read(),q = read();
     for(re int i=1; i<=n; i++)
     { x[i] = read(); }
     for(re int i=1; i<=q; i++)
     { len[i] = read(); }
     if(q==0)
     { return 0; }
     if(n<=1e3)
     { task1(); }
     else
     {
       task2();
       //task3();
       //task4();
     }
     return 0;
   }
}
signed main()
{ return OMA::main(); }

正解不会...，所以，



所以为什么给的std只能拿30pts啊，单调还判错了

updated on 7.21

好吧，std换成对的了 是指针_{好吧，是迭代器}，可恶

没改出来，贴下std吧..

std

#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
const int maxn=1e5+10;
int n,m;
ll tot,ans[maxn];
vector<int> x;
vector<pair<int,int> > a;
map<int,int> mp;

inline ll calc(ll k){
	if(!mp.empty()&&mp.begin()->second<0)
		return tot-(mp.size()-1)*k;
	else
		return tot-mp.size()*k;
}
inline void solve(){
	priority_queue<pair<int,int>,vector<pair<int,int> >,greater<pair<int,int> > > q;
	int t=0;
	for(int i=0;i<x.size();++i){
		tot+=abs(x[i]);
		mp[i]=x[i];
		q.push(make_pair(abs(x[i]),i));
	}
	while(!q.empty()){
		int id=q.top().second,tmp=q.top().first;q.pop();
		map<int,int>::iterator p=mp.lower_bound(id);
		if(p==mp.end()||p->first!=id||abs(p->second)!=tmp)
			continue;
		while(t<a.size()&&abs(p->second)>a[t].first)
			ans[a[t].second]=calc(a[t].first),++t;
		if(p!=mp.begin())
			if(p!=prev(mp.end())){
				tmp=p->second,tot-=abs(p->second);
				tmp+=prev(p)->second,tot-=abs(prev(p)->second);
				tmp+=next(p)->second,tot-=abs(next(p)->second);
				mp.erase(prev(p));
				mp.erase(next(p));
				p->second=tmp,tot+=abs(tmp);
				q.push(make_pair(abs(tmp),id));
			}
			else{
				tot-=abs(p->second);
				mp.erase(p);
			}
		else if(p->second>0)
			if(p!=prev(mp.end())){
				tmp=p->second,tot-=abs(p->second);
				tmp+=next(p)->second,tot-=abs(next(p)->second);
				mp.erase(next(p));
				if(tmp){
					p->second=tmp,tot+=abs(tmp);
					q.push(make_pair(abs(tmp),id));
				}
				else
					mp.erase(p);
			}
			else{
				tot-=abs(p->second);
				mp.erase(p);
			}
	}
	while(t<a.size())
		ans[a[t].second]=calc(a[t].first),++t;
}

int main(){

	scanf("%d%d",&n,&m);
	for(int i=0,p,last=0;i<n;++i){
		scanf("%d",&p);
		if(p==last)
			continue;
		if(!x.empty()&&(x.back()<0&&p<last||x.back()>0&&p>last))
			x.back()+=p-last;
		else
			x.push_back(p-last);
		last=p;
	}
	for(int i=0,l;i<m;++i){
		scanf("%d",&l);
		a.push_back(make_pair(l,i));
	}
	sort(a.begin(),a.end());
	solve();
	for(int i=0;i<m;++i)
		printf("%lld\n",ans[i]);
	return 0;
}

所以 \(\frac{1}{4}\) 是有多喜欢STL啊