The longest common sub-sequence problem of LSC

Same as most LCS algorithms on the web. Implementations are implemented using DP. The state transfer equation is not written. Skilled in the heart anyway.

Prints the shortest sequence that contains a and B. Of course, the relative position of the characters in A and b in the new sequence will not change.

The main is to familiarize yourself with the code:

#include <cstdio>#include<cstring>using namespacestd;Chara[ +],b[ +];intvisit[ +][ +],dp[ +][ +];voidpriintXinty) {    if(x==0&&y==0)        return ; if(visit[x][y]==1) {pri (x-1, Y1); printf ("%c", a[x-1]); }Else if(visit[x][y]==2) {pri (x-1, y); printf ("%c", a[x-1]); }Else if(visit[x][y]==3) {pri (x, y-1); printf ("%c", b[y-1]); }Else if(visit[x][y]==0&&x>0&&y==0) {pri (x-1, y); printf ("%c", a[x-1]); }Else if(visit[x][y]==0&&y>0&&x==0) {pri (x, y-1); printf ("%c", b[y-1]); }}intMain () { while(SCANF ("%s%s", B)) {        intLa=strlen (a), lb=strlen (b); Memset (DP,0,sizeof(DP)); memset (Visit,0,sizeof(visit));  for(intI=1; i<=la;i++)        {             for(intj=1; j<=lb;j++)            {                if(a[i-1]==b[j-1]) {Dp[i][j]=dp[i-1][j-1]+1; VISIT[I][J]=1; }                Else if(dp[i-1][j]>dp[i][j-1]) {Dp[i][j]=dp[i-1][j]; VISIT[I][J]=2; }                Else if(dp[i-1][j]<=dp[i][j-1]) {Dp[i][j]=dp[i][j-1]; VISIT[I][J]=3;        }}} pri (LA,LB); printf ("\ n"); }    return 0;}

More familiar with the code. It is certainly best to be clear about the heart. Come on.

The longest common sub-sequence problem of LSC