hdu 1159, LCS, dynamic programming, recursive backtrack vs iterative backtrack vs incremental, C++

thanks prof. Abhiram Ranade for his vedio on Longest Common Subsequence ‘s back track search view in lecture 19, nice explanation indeed.

// back track, recursive, 390 ms, O(m*n) memory

#include 
#include 
#include 
#include 
#include 

#define MAXSIZE 1001
using std::string;


char *p1, *p2;
int M, N;
int table[MAXSIZE][MAXSIZE];

int solveLCSlength(int i, int j) {
    if(table[i][j]>=0) return table[i][j];
    int k;
    for(k=j;kif(k!=N) table[i][j]=std::max(solveLCSlength(i+1,j),1+solveLCSlength(i+1,k+1));
    else table[i][j]=solveLCSlength(i+1,j);
    return table[i][j];
}

int LCSlength(string &s1, string &s2) {
    p1=&s1[0], p2=&s2[0], M=s1.size(), N=s2.size();
    for(int i=0;ifor(int j=0;j1; }
        table[i][N]=0;
    }
    for(int j=0;j<=N;++j) { table[M][j]=0; }
    return solveLCSlength(0,0);
}

int main() {
    string s1, s2;
    while(std::cin >> s1 >> s2) {
        printf("%d\n",LCSlength(s1,s2));
    }
    return 0;
}

// turn recursive backtrack version to iterative and reduce memory cost to O(m+n) , 312ms

#include 
#include 
#include 

#define MAXSIZE 1001

int LCSlength(char *s1, char *s2) {
    static int table[MAXSIZE<<1];
    int len1,len2, i,j,k, *prev,*curr;
    len1=strlen(s1), len2=strlen(s2);
    prev=&table[0], curr=&table[len2+1];
    for(i=0;i<=len2;++i) prev[i]=0;
    curr[len2]=0;
    for(i=len1-1;i>=0;--i) {
            char tmp=s1[i];
        for(j=len2-1;j>=0;--j) {
            for(k=j;kif(k!=len2 && curr[j]==prev[k+1]) ++curr[j];
        }
        std::swap(prev,curr);
    }
    return prev[0];
}

int main() {
    char s1[MAXSIZE], s2[MAXSIZE];
    while(scanf("%s%s",s1,s2)==2) {
        printf("%d\n",LCSlength(s1,s2));
    }
    return 0;
}

// incremental, 31ms, O(m+n) memory

#include 
#include 

#define MAXSIZE 1001

int main() {
    char s1[MAXSIZE], s2[MAXSIZE];
    int table[MAXSIZE<<1], *prev, *curr;
    int len1,len2, i,j;
    while(scanf("%s%s",s1,s2)==2) {
        len1=strlen(s1), len2=strlen(s2);
        prev=table, curr=&table[len2+1];
        for(i=0;i<=len2+1;++i) prev[i]=0;
        for(i=1;i<=len1;++i) {
            for(j=1;j<=len2;++j) {
                if(s1[i-1]==s2[j-1]) curr[j]=1+prev[j-1];
                else curr[j]=prev[j]>curr[j-1]?prev[j]:curr[j-1];
            }
            std::swap(curr,prev);
        }
        printf("%d\n",prev[len2]);
    }
    return 0;
}