【TJOI2019 Day2】简要题解

T1：

传送门
傻逼题
考虑到$k$很小，把$2k+1$种暴力建出来跑$dij$就可以了
结果出题人竟然不卡优先队列的$O(n+mklog)$写法
跑的还比$set的O(m+nklog)$快……

#include<bits/stdc++.h>
using namespace std;
const int RLEN=1<<20|1;
inline char gc(){
    static char ibuf[RLEN],*ib,*ob;
    (ob==ib)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
    return (ob==ib)?EOF:*ib++;
}
#define gc getchar
inline int read(){
    char ch=gc();
    int res=0,f=1;
    while(!isdigit(ch))f^=ch=='-',ch=gc();
    while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
    return f?res:-res;
}
#define ll long long
#define re register
#define pii pair<int,int>
#define fi first
#define se second
#define pb push_back
#define cs const
#define bg begin
inline void chemx(int &a,int b){a<b?a=b:0;}
inline void chemn(int &a,int b){a>b?a=b:0;}
cs int N=300005;
vector<pii> e[N];
int n,m,k,dis[N],vis[N],val[N],id[10005][23],tot,str,des;
inline void addedge(int u,int v,int w){
	for(int i=1;i<=2*k+1;i++)
	if(id[v][i+val[v]])e[id[u][i]].pb(pii(id[v][i+val[v]],w));
}
priority_queue<pii,vector<pii>,greater<pii> >q;
inline int dijkstra(){
	memset(dis,127,sizeof(int)*(tot+1));
	memset(vis,0,sizeof(int)*(tot+1));
	int s=id[str][k+1+val[str]];
	dis[s]=0,q.push(pii(0,s));
	while(!q.empty()){
		int u=q.top().se;q.pop();
		if(vis[u])continue;
		vis[u]=1;
		for(pii &x:e[u]){
			if(dis[x.fi]>dis[u]+x.se){
				dis[x.fi]=dis[u]+x.se;
				q.push(pii(dis[x.fi],x.fi));
			}
		}
	}
	int res=dis[0];
	for(int i=1;i<=2*k+1;i++)chemn(res,dis[id[des][i]]);
	return res==dis[0]?-1:res;
}
inline void solve(){
	n=read(),m=read(),k=read(),tot=0;
	for(int i=1,l=2*k+1;i<=n;i++)for(int j=1;j<=l;j++)id[i][j]=++tot;
	for(int i=1;i<=n;i++)val[i]=read()==2?1:-1;
	for(int i=1;i<=m;i++){
		int u=read(),v=read(),w=read();
		addedge(u,v,w),addedge(v,u,w);
	}
	str=read(),des=read();
	cout<<dijkstra()<<'
';
	for(int i=1;i<=tot;i++){
		e[i].clear();
	}
	for(int i=1;i<=n;i++)
	for(int j=1;j<=2*k+1;j++)id[i][j]=0;
}
int main(){
	int T=read();
	while(T--)
		solve();
}

T2：

传送门
傻逼题
把$Sam$建出来取$right$集合大小为$k$的
做区间加可以写前缀和

#include<bits/stdc++.h>
using namespace std;
const int RLEN=1<<20|1;
inline char gc(){
    static char ibuf[RLEN],*ib,*ob;
    (ob==ib)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
    return (ob==ib)?EOF:*ib++;
}
#define gc getchar
inline int read(){
    char ch=gc();
    int res=0,f=1;
    while(!isdigit(ch))f^=ch=='-',ch=gc();
    while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
    return f?res:-res;
}
#define ll long long
#define re register
#define pii pair<int,int>
#define fi first
#define se second
#define pb push_back
#define cs const
#define bg begin
inline void chemx(int &a,int b){a<b?a=b:0;}
inline void chemn(ll &a,ll b){a>b?a=b:0;}
cs int N=200005;
int tt1,tt2;

namespace Sam{
	int nxt[N][26],fail[N],len[N],siz[N],tot,last;
	inline void init(){tot=last=1;}
	inline void insert(int c){
		int cur=++tot,p=last;last=tot;
		len[cur]=len[p]+1,siz[cur]=1;
		for(;p&&!nxt[p][c];p=fail[p])nxt[p][c]=cur;
		if(!p)fail[cur]=1;
		else{
			int q=nxt[p][c];
			if(len[p]+1==len[q])fail[cur]=q;
			else{
				int clo=++tot;
				memcpy(nxt[clo],nxt[q],sizeof(nxt[q]));
				len[clo]=len[p]+1,fail[clo]=fail[q];
				for(;p&&nxt[p][c]==q;p=fail[p])nxt[p][c]=clo;
				fail[q]=fail[cur]=clo;
			}
		}
	}
	int a[N],b[N],buc[N],tag[N];
	inline void buc_sort(int k){
		for(int i=1;i<=tot;i++)a[len[i]]++;
		for(int i=1;i<=tot;i++)a[i]+=a[i-1];
		for(int i=tot;i;i--)b[a[len[i]]--]=i;
		for(int i=tot;i;i--)siz[fail[b[i]]]+=siz[b[i]];
		for(int i=1;i<=tot;i++){
			if(siz[i]==k)
			tag[len[i]+1]--,tag[len[fail[i]]+1]++;
		}
		for(int i=1;i<=tot;i++)buc[i]=buc[i-1]+tag[i];
		int mx=0;
		for(int i=tot;i;i--){
			if(buc[i]>buc[mx])mx=i;
		}
		if(mx)cout<<mx<<'
';
		else cout<<-1<<'
';
	}
	inline void clear(){
		for(int i=1;i<=tot;i++){
			memset(nxt[i],0,sizeof(nxt[i]));
			fail[i]=len[i]=siz[i]=a[i]=b[i]=buc[i]=tag[i]=0;
		}
	}
}
char s[N];
int k;
int main(){
	int T=read();
	while(T--){
		scanf("%s",s+1);
		Sam::init();
		for(int i=1,len=strlen(s+1);i<=len;i++)Sam::insert(s[i]-'a');
		k=read(),Sam::buc_sort(k),Sam::clear();
	}
}

T3：

传送门

其实是原题，$CF 750G$
题解传送门

第一问直接暴力跳就可以了
不过这里限制了树高
需要判一下（调了一万年）

#include<bits/stdc++.h>
using namespace std;
const int RLEN=1<<20|1;
inline char gc(){
    static char ibuf[RLEN],*ib,*ob;
    (ob==ib)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
    return (ob==ib)?EOF:*ib++;
}
#define gc getchar
inline int read(){
    char ch=gc();
    int res=0,f=1;
    while(!isdigit(ch))f^=ch=='-',ch=gc();
    while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
    return f?res:-res;
}
#define ll long long
inline ll readl(){
    char ch=gc();
    ll res=0,f=1;
    while(!isdigit(ch))f^=ch=='-',ch=gc();
    while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
    return f?res:-res;
}
#define re register
#define pii pair<int,int>
#define fi first
#define se second
#define pb push_back
#define cs const
#define bg begin
cs int N=55;
int d,c;
ll a,b,bin[N];
inline ll solve1(ll x,ll y){
	ll res=0;
	while(x!=y){
		if(x<y)swap(x,y);
		res+=x,x>>=1;
	}
	res+=x;return res;
}
ll f[N][N*2][2];
inline ll dp(ll mx,int l,int r,int t){
	memset(f,0,sizeof(f));
	f[0][0][0]=1;int len=1;
	for(int i=1;bin[i-1]<=mx;i++,len++){
		int d=(mx>>i)&1;
		for(int j=0;j<=t;j++){
			for(int x=0;x<=1;x++)
			if(x!=1||i<l)
			for(int y=0;y<=1;y++)
			if(y!=1||i<r){
				if((x+y)%2==d)f[i][j+x+y][(x+y)/2]+=f[i-1][j][0];
				else f[i][j+x+y][(x+y+1)/2]+=f[i-1][j][1];
			}
		}
	}
	return f[len-1][t][0];
}
inline ll solve2(ll s){
	ll ans=0;
	for(int i=0;i<d&&bin[i]<=s;i++)
	for(int j=0;j<d&&bin[j]<=s;j++){
		ll x=(s-bin[j]+1)/(bin[i+1]+bin[j+1]-3);
		if(max(i,j)+1+(int)log2(x)>d)continue;
		if(!x)continue;
		ll res=s-bin[j]+1-(bin[i+1]+bin[j+1]-3)*x;
		if(!res){ans++;continue;}
		for(int k=1;k<=i+j;k++)
		if((res+k)%2==0)ans+=(dp(res+k,i,j,k));
	}
	return ans-1;
}
inline void solve(){
	d=read(),a=readl(),b=readl(),c=read();
	ll now=solve1(a,b);
	if(c==1)cout<<now<<'
';
	if(c==2)cout<<solve2(now)<<'
';
}
int main(){
	for(int i=0;i<N;i++)bin[i]=1ll<<i;
	int T=read();
	while(T--)solve();
}