首先基礎dp方程很好寫:
1.若a[i]==b[j] dp[i][j]=min(dp[i-1][j],dp[i][j-1])+1
2.若a[i]!=b[j] dp[i][j]=dp[i-1][j-1]
時間複雜度
而dp[][]二維數組,很明顯空間複雜度過大,於是思考跳的過程中,對於a[i]!=b[j],i,j能更新的狀態是唯一的,即dp[i+1]j和dp[i]j+1,而若括號裏的任意一種條件都不滿足,則這種狀態永遠都不會被搜索到,於是不記錄a[i]!=b[j]的狀態,僅用一個一維數組記錄a[i]==b[j]的狀態即可。
ac代碼
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
const int maxn=1000005;
int n;
int a[maxn],b[maxn],pos[maxn],dp[maxn];
vector<int>s[maxn<<1];
int read()
{
int x=0;char ch=getchar();
while(ch<'0'||ch>'9')ch=getchar();
while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
return x;
}
int memorial_search(int x,int y)
{
if(!x||!y) return x+y;
if(a[x]==b[y])
{
if(dp[x]!=-1)return dp[x];
return dp[x]=min(memorial_search(x-1,y),memorial_search(x,y-1))+1;
}
else
{
int now=lower_bound(s[x-y+n].begin(),s[x-y+n].end(),x)-s[x-y+n].begin()-1;
if(now<0)
return max(x,y);
else
{
int delta=x-s[x-y+n][now];
return memorial_search(x-delta,y-delta)+delta;
}
}
}
int main()
{
scanf("%d",&n);
for(int i=1;i<=n;i++)a[i]=read();
for(int i=1;i<=n;i++)pos[b[i]=read()]=i;
memset(dp,-1,sizeof dp);
for(int i=1;i<=n;i++)
{
int delta=i-pos[a[i]]+n;//+n防溢出邊界
s[delta].push_back(i);
}
printf("%d",memorial_search(n,n));
return 0;
}
原始代碼
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
const int maxn=5005;
int n;
int a[maxn],b[maxn],pos[maxn],dp[maxn][maxn];
vector<int>s[maxn<<1];
int read()
{
int x=0;char ch=getchar();
while(ch<'0'||ch>'9')ch=getchar();
while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
return x;
}
int memorial_search(int x,int y)
{
if(dp[x][y]!=-1)return dp[x][y];
if(!x||!y) return x+y;
int &res=dp[x][y];
if(a[x]==b[y])res=min(memorial_search(x-1,y),memorial_search(x,y-1))+1;
else
{
int now=lower_bound(s[x-y+n].begin(),s[x-y+n].end(),x)-s[x-y+n].begin()-1;
if(now<0)
{
if(x>=y)res=memorial_search(x-y,0)+y;
else res=memorial_search(0,y-x)+x;
}
else
{
int delta=x-s[x-y+n][now]+1;
res=memorial_search(x-delta,y-delta)+delta+1;
}
}
return res;
}
int main()
{
scanf("%d",&n);
for(int i=1;i<=n;i++)a[i]=read();
for(int i=1;i<=n;i++)pos[b[i]=read()]=i;
memset(dp,-1,sizeof dp);
for(int i=1;i<=n;i++)
{
int delta=i-pos[a[i]]+n;//+n防溢出邊界
s[delta].push_back(i);
}
printf("%d",memorial_search(n,n));
return 0;
}
GET:
沒事先寫個基礎dp,說不定優化搞着搞着就搞出來了。。。