[APIO2014]

T1.回文树裸题。

#include<cstdio>
#include<iostream>
#define ll long long
using namespace std;
inline int read()
{
    int x=,f=;char ch=getchar();
    while(ch<''||ch>''){if(ch=='-') f=-;ch=getchar();}
    while(ch>=''&&ch<=''){x=x*+ch-'';}
    return x*f;
}

int cnt=;
int f[],l[],p[];
int c[][];
char s[];

int main()
{
    freopen("palindrome.in","r",stdin);
    freopen("palindrome.out","w",stdout);
    scanf("%s",s+);
    f[]=;l[++cnt]=-;
    for(int i=,j=;s[i];++i)
    {
        while(s[i]!=s[i-l[j]-]) j=f[j];
        if(!c[j][s[i]-'a'])
        {
            l[++cnt]=l[j]+;
            int k=f[j];
            while(s[i]!=s[i-l[k]-])
                k=f[k];
            f[cnt]=c[k][s[i]-'a'];
            c[j][s[i]-'a']=cnt;
        }
        j=c[j][s[i]-'a'];
        ++p[j];
    }
    ll ans=;
    for(int i=cnt;i>;i--)
    {
        ans=max(ans,1ll*p[i]*l[i]);
        p[f[i]]+=p[i];
    }
    cout<<ans;
    return ;
}

T2.斜率优化

f[i]=max(f[j]+s[j]*(s[i]-s[j]))

令g[i]=f[i]-s[i]^2

j比k优 那么g[j]+s[i]s[j]>g[k]+s[i][k]

g[j]-g[k]>s[i](s[k]-s[j])

g[j]-g[k]/(s[j]-s[k])<-s[i]

所以维护队列上凸即可

#include<iostream>
#include<cstdio>
#include<queue>
#define MAXN 100000
#include<cstring>
#define ll long long
using namespace std;
int read()
{
    int x=,f=;char ch=getchar();
    while(ch<''||ch>''){if(ch=='-') f=-;ch=getchar();}
    while(ch>=''&&ch<=''){x=x*+ch-''; ch=getchar();}
    return x*f;
} 

int n,k;
int from[][MAXN+];
ll f[MAXN+];
ll s[MAXN+];
ll g[][MAXN+];
int q[MAXN+];
int top=,tail=;

ll get(int pre,int x,int t)
{
    while(top-tail>=&&(g[pre][q[tail]]-g[pre][q[tail+]])<s[x]*(s[q[tail+]]-s[q[tail]]))
    ++tail;
    from[t][x]=q[tail];
    return g[pre][q[tail]]+s[q[tail]]*s[x];
};

void ins(int t,int k)
{
    while(top-tail>=)
    {
        int i=q[top-],j=q[top];
        if((g[t][k]-g[t][j])*(s[i]-s[j])<(g[t][j]-g[t][i])*(s[j]-s[k])) top--;
        else break;
    }
    q[++top]=k;
}

int main()
{
    //freopen("sequence.in","r",stdin);
    //freopen("sequence.out","w",stdout);
    n=read();k=read();
    for(int i=;i<=n;i++)
    {
        s[i]=read();s[i]+=s[i-];
    }
    int pre=,nown=;
    for(int i=;i<=n;i++) g[pre][i]=-(s[i]*s[i]);
    for(int l=;l<=k+;l++)
    {
        memset(f,,sizeof(f));top=;tail=;
        memset(g[nown],,sizeof(g[nown]));
        ins(pre,l-);
        for(int i=l;i<=n;i++)
        {
            f[i]=get(pre,i,l);
            g[nown][i]=f[i]-(s[i]*s[i]);
            if(s[i]!=s[i-])ins(pre,i);
            //printf("%d %d %d %I64d\n",l,i,from[l][i],f[i]);
        }
        nown=-nown;
        pre=-pre;
    }
    printf("%lld\n",f[n]);
    return ;
}

T3.

树形dp，把根节点作为第一个存在的点，那么每一段蓝线总是 爷爷-爸爸-点  的形式。

所以用f1表示一个点作为中间的点的最大值，f2表示一个点不是中间点的最大答案。

树形dp之后O(1)换根

#include<iostream>
#include<cstdio>
#include<cstring>
#define ll long long
#define MAXN 200000
#define INF 2000000000
using namespace std;
int read()
{
    int x=,f=;char ch=getchar();
    while(ch<''||ch>''){if(ch=='-') f=-;ch=getchar();}
    while(ch>=''&&ch<=''){x=x*+ch-''; ch=getchar();}
    return x*f;
} 

ll ans=;
int n,cnt=;
ll f[MAXN+],f2[MAXN+];
int mx[MAXN+],mx2[MAXN+],from[MAXN+],from2[MAXN+];
int fa[MAXN+],head[MAXN+],p[MAXN+];

struct edge{
    int to,next,w;
}e[MAXN*+];

void ins(int f,int t,int w)
{
    e[++cnt].next=head[f];
    head[f]=cnt;
    e[cnt].to=t;
    e[cnt].w=w;
}

void renew(int x,int t,int f)
{
    //cout<<"ins"<<x<<" "<<t<<" "<<f<<endl;
    if(t>mx[x])
    {
        from2[x]=from[x];
        mx2[x]=mx[x];
        mx[x]=t;
        from[x]=f;
    }
    else if(t>mx2[x])
    {
        mx2[x]=t;
        from2[x]=f;
    }
}

void dfs(int x,int pre,int father)
{
    fa[x]=father;
    for(int i=head[x];i;i=e[i].next)
    {
        int v=e[i].to;
        if(v!=fa[x])
        {
            dfs(v,e[i].w,x);f2[x]+=max(f[v],f2[v]);
            renew(x,e[i].w+f2[v]-max(f[v],f2[v]),v);
        }
    }
    f[x]=f2[x]+mx[x]+pre;
    if(f[x]<) f[x]=;
    //cout<<x<<" "<<f[x]<<" "<<f2[x]<<endl;
}

void solve(int x)
{
    if(fa[x])
    {
        int v=fa[x];f2[x]+=max(f[v],f2[v]);
        renew(x,p[x]+f2[v]-max(f[v],f2[v]),v);
    }
    ans=max(ans,f2[x]);f[x]=f2[x]+mx[x];
    //cout<<"solve"<<x<<" "<<f2[x]<<" "<<f[x]<<endl;
    for(int i=head[x];i;i=e[i].next)
    {
        int v=e[i].to;
        if(v!=fa[x])
        {
            ll t2=max(f[v],f2[v]);
            f[x]+=e[i].w-t2;
            if(from[x]==v) f[x]=f[x]-mx[x]+mx2[x];
            f2[x]-=t2;p[v]=e[i].w;
            solve(v);
            f2[x]+=t2;
            f[x]=f2[x]+mx[x];
        }
    }
}

int main()
{
    freopen("beads.in","r",stdin);
    freopen("beads.out","w",stdout);
    n=read();
    for(int i=;i<n;i++)
    {
        int u=read(),v=read(),w=read();
        ins(u,v,w);ins(v,u,w);
    }
    for(int i=;i<=n;i++) mx[i]=mx2[i]=-INF;
    dfs(,,);
    solve();
    cout<<ans;
    return ;
}