$ZOJ\ 2432\ Greatest\ Common\ Increasing\ Subsequence$

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$Description$

求两个序列的最长公共上升子序列

$Solution$

$f[i][j]$表示$a$序列匹配到$i$和$b$序列匹配到$j$的最长上升序列的长度，这里并不要求$a[i]==b[j]$。

两层循环，外层$i=1...n1$,内层$j=1...n2$

转移：

$f[i][j]=max(f[i-1][1...j-1](if\  b[j]<a[i]))+1$

这里的$max(f[i-1][i...j-1](if\  b[j]<a[i]))$不用每次单独枚举，只要在循环内即使更新就可以了

或者,也可以用树状数组来维护.

$Code$

#include<bits/stdc++.h>
#define Ri register int
#define il inline
#define go(i,a,b) for(Ri i=a;i<=b;++i)
#define yes(i,a,b) for(Ri i=a;i>=b;--i)
#define e(i,u) for(Ri i=b[u];i;i=a[i].nt)
#define ll long long
#define db double
using namespace std;
il int read()
{
    Ri x=,y=;char c=getchar();
    while(c<'' || c>''){if(c=='-')y=-;c=getchar();}
    while(c>='' && c<=''){x=(x<<)+(x<<)+c-'';c=getchar();}
    return x*y;
}
const int N=;
int n,m,a[N],b[N],f[N][N],as,d[N],t[N][N*];
struct nd{int d,p;}c[N];
il bool cmp(nd x,nd y){return x.d<y.d;}
il int lowbit(Ri x){return x&(-x);}
il void change(Ri i,Ri j,Ri dat)
{
    while(i<=n)
    {
        t[j][i]=max(t[j][i],dat);
        i+=lowbit(i);
    }
}
il int sum(Ri i,Ri j)
{
    Ri ret=;
    while(i>)
    {
        ret=max(ret,t[j][i]);
        i-=lowbit(i);
    }
    return ret;
}
int main()
{
    freopen("lcis.in","r",stdin);
    freopen("lcis.out","w",stdout);
    n=read(),m=read();
    go(i,,n)a[i]=read(),c[i]=(nd){a[i],i};
    go(i,,m)b[i]=read();
    sort(c+,c+n+,cmp);
    go(i,,n)
        if(i>= && c[i].d==c[i-].d)d[c[i].p]=d[c[i-].p];
        else d[c[i].p]=i;
    go(i,,n)
    {
        go(j,,m)
        {
            f[i][j]=max(f[i][j],f[i][j-]);
            if(a[i]==b[j])
                f[i][j]=max(f[i][j],sum(d[i]-,j-)+);
        }
        go(j,,m)change(d[i],j,f[i][j]);
    }
    go(i,,n)as=max(as,f[i][m]);
    printf("%d\n",as);
    return ;
}

View 树状数组 Code

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #define Rg register
 #define il inline
 #define db double
 #define ll long long
 #define mem(a,b) memset(a,b,sizeof(a));
 #define go(i,a,b) for(Rg int i=a;i<=b;i++)
 #define yes(i,a,b) for(Rg int i=a;i>=b;i--)
 using namespace std;
 il int read()
 {
     int x=,y=;char c=getchar();
     while(c<''||c>''){if(c=='-')y=-;c=getchar();}
     while(c>=''&&c<=''){x=(x<<)+(x<<)+c-'';c=getchar();}
     return x*y;
 }
 const int N=;
 int T,n1,n2,ans,a[N],b[N],al[N],f[N][N],lst[N][N];
 il void sol()
 {
     ans=;mem(f,);int x=n1,y=,t;
     go(i,,n1)
     {
         t=;
         go(j,,n2)
         {
             if(a[i]!=b[j])f[i][j]=f[i-][j];
             if(a[i]>b[j]&&f[i][j]>f[i][t])t=j;
             if(a[i]==b[j])f[i][j]=f[i][t]+,lst[i][j]=t;
             //cout<<"f["<<i<<"]["<<j<<"]:"<<f[i][j]<<endl;
         }
     }
     go(i,,n2)if(f[n1][i]>ans)ans=f[n1][i],y=i;
     t=;
     while(f[x][y])
     {
         while(a[x]!=b[y]){x--;if(!x)return;}
         al[ans-t]=b[y],y=lst[x][y],t++;
     }
 }
 int main()
 {
     T=read();
     while(T--)
     {
         n1=read();go(i,,n1)a[i]=read();
         n2=read();go(i,,n2)b[i]=read();
         sol();
         printf("%d\n",ans);
         if(ans){go(i,,ans)printf("%d ",al[i]);printf("\n");}
     }
     return ;
 }