topic
Dongdong has two sequences A and B.
He wants to know the length of the LIS of sequence A and the LCS of sequence AB.
Note that LIS is strictly increasing, that is, a1<a2<...<ak(ai<=1,000,000,000).
Input
Two numbers in the first line n, m (1<=n<=5,000,1<=m<=5,000)
The number of n in the second line represents the sequence A
The number of m in the third line represents the sequence B
Output
Output a row of data ans1 and ans2, which represent the length of the LIS of sequence A and the LCS of sequence AB, respectively
Simple Input
5 5
1 3 2 5 4
2 4 3 1 5
Simple Output
3 2
Ideas
error
1. Forgot to enter n and m.
2. You don’t need to use recursion for this question, just use nested for loop, because when calculating f[i][j], the previous ones of f[i][j] have already been calculated
Code
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;
//可能需要二分orz
const int maxn=5010;
long long a[maxn];
long long b[maxn];
int f[maxn];
int f2[maxn][maxn]={
};
int main()
{
int n,m,ans1=0;
cin>>n>>m;
for(int i=1;i<=n;i++)
scanf("%lld",&a[i]);
for(int i=1;i<=m;i++)
scanf("%lld",&b[i]);
//LIS 最长上升子序列
f[1]=1;
for(int i=2;i<=n;i++)
{
int maxx=0;
for(int j=1;j<i;j++)
{
if(a[j]<a[i]&&f[j]>maxx)
maxx=f[j];
}
f[i]=maxx+1;
}
for(int i=1;i<=n;i++)
if(f[i]>ans1)
ans1=f[i];
cout<<ans1<<' ';
f2[1][0]=0;
f2[0][1]=0;
f2[0][0]=0;
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
{
if(a[i]==b[j])
f2[i][j]=f2[i-1][j-1]+1;
else
f2[i][j]=max(f2[i-1][j],f2[i][j-1]);
}
cout<<f2[n][m]<<endl;
return 0;
}