_ALIVE_LYY @ 2018-02-06 23:29:45
using namespace std;
int a[1005],b[1005];
int f[1005][1005];
int main()
{
int n;
cin>>n;
for(int i=1;i<=n;i++){
cin>>a[i];
}
for(int i=1;i<=n;i++){
cin>>b[i];
}
f[0][0]=0;
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++){
if(a[i]==b[j]){
f[i][j]=f[i-1][j-1]+1;
}
else {
f[i][j]=max(f[i][j-1],f[i-1][j]);
}
}
}
cout<<f[n][n];
}
by wucstdio @ 2018-03-18 15:24:15
你这是O(n^2)的做法,只能过1000以内的,当然RE了,要用nlogn的做法
by 做梦都想AK @ 2018-08-24 15:36:35
@wucstdio 话说大佬nlogn的做法咋写
by Clearlove7loveyou @ 2018-08-24 15:37:09
@qq770529510 +1