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最新推荐文章于 2023-12-07 13:14:17 发布

原创最新推荐文章于 2023-12-07 13:14:17 发布 · 533 阅读

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本文介绍了一种通过字符串处理和哈希技术来判断两个DNA序列是否相似的方法。使用暴力枚举和折半搜索策略，结合自定义哈希函数对DNA序列进行编码，最终通过集合运算确定两序列的相似度。

D DNA Analysis

暴力枚举第一次翻转，记正反串。
折半搜索

#include<bits/stdc++.h> 
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,0x3f,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MEMx(a,b) memset(a,b,sizeof(a));
#define INF (0x3f3f3f3f)
#define F (1000000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int> 
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;
#define PRi2D(a,n,m) For(i,n) { \
                        For(j,m-1) cout<<a[i][j]<<' ';\
                        cout<<a[i][m]<<endl; \
                        } 
#pragma comment(linker, "/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
#define gmax(a,b) a=max(a,b);
#define gmin(a,b) a=min(a,b);
#define fr(x) freopen(x,"r",stdin)
#define fw(x) freopen(x,"w",stdout)
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
inline int read()
{
    int x=0,f=1; char ch=getchar();
    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
    return x*f;
} 
#define MAXN (40)
string Filename="dna";
char s[40],s2[40];
int n;
ull seed=123;
ull f[MAXN][MAXN],g[MAXN][MAXN],p2[MAXN];
set<ull> S;
void get() {
    cin>>(s+1);
    p2[0]=1;
    For(i,35) p2[i]=p2[i-1]*seed;
    int n=strlen(s+1);
    For(i,n) {
        Fork(j,i,n) {
            reverse(s+i,s+j+1);     
            For(i,n) {
                f[i][i]=g[i][i]=s[i];
            }
            Fork(len,2,n) {
                For(i,n-len+1) {
                    int j=i+len-1;
                    f[i][j]=f[i][j-1]*seed+s[j];
                    g[i][j]=g[i+1][j]*seed+s[i];
                }
            }

            For(k,n)
                Fork(l,k,n) {
                    ll ans=0;
                    if (1<k) ans=f[1][k-1];
                    ans=ans*p2[l-k+1]+g[k][l];
                    if (l<n) ans=ans*p2[n-l]+f[l+1][n];
//                  cout<<ans<<' ';
                    S.insert(ans);
                }
            reverse(s+i,s+j+1);     

        }

    }
} 
bool get2() {
    cin>>(s+1);
    p2[0]=1;
    For(i,35) p2[i]=p2[i-1]*seed;
    int n=strlen(s+1);
    For(i,n) {
        Fork(j,i,n) {
            reverse(s+i,s+j+1);     
            For(i,n) {
                f[i][i]=g[i][i]=s[i];
            }
            Fork(len,2,n) {
                For(i,n-len+1) {
                    int j=i+len-1;
                    f[i][j]=f[i][j-1]*seed+s[j];
                    g[i][j]=g[i+1][j]*seed+s[i];
                }
            }

            For(k,n)
                Fork(l,k,n) {
                    ll ans=0;
                    if (1<k) ans=f[1][k-1];
                    ans=ans*p2[l-k+1]+g[k][l];
                    if (l<n) ans=ans*p2[n-l]+f[l+1][n];
                    if (S.count(ans)) return 1;
                }
            reverse(s+i,s+j+1);     

        }

    }
    return 0;
} 
int main()
{
    freopen((Filename+".in").c_str(), "r", stdin);
    freopen((Filename+".out").c_str(), "w", stdout);
    get();
    puts(get2() ? "Similar": "Different")  ;
    return 0;
}