CF463D Gargari and Permutations dp

Posted 2020-12-01 guangheli

考试 T2，推一推发现可以转换成一个 dp 模型. 

code: 

#include <bits/stdc++.h>      
#define N 1004  
using namespace std; 
void setIO(string s) 
{
    string in=s+".in";   
    string out=s+".out";   
    freopen(in.c_str(),"r",stdin);   
    freopen(out.c_str(),"w",stdout);   
}   
struct node 
{ 
    int l,r;   
}t[N];               
int a[12][N],pos[N],f[N],L[12][N];    
int main() 
{ 
    // setIO("seq"); 
    int n,i,j,m,ans=1;    
    scanf("%d%d",&n,&m);   
    for(i=1;i<=m;++i)   for(j=1;j<=n;++j)   scanf("%d",&a[i][j]);                  
    for(i=1;i<=n;++i)   pos[a[1][i]]=i;    
    for(i=1;i<=m;++i)   for(j=1;j<=n;++j)   a[i][j]=pos[a[i][j]];   
    for(i=1;i<=m;++i)   for(j=1;j<=n;++j)   L[i][a[i][j]]=j;   
    f[1]=1;   
    for(i=2;i<=n;++i) 
    {      
        f[i]=1;  
        for(j=1;j<i;++j)  
        { 
            int flag=0;  
            for(int k=1;k<=10;++k)    if(L[k][j]>L[k][i])   flag=1;   
            if(!flag)    f[i]=max(f[i], f[j]+1),  ans=max(ans, f[i]);     
        }     
    } 
    printf("%d
",ans);   
    return 0;    
}