动态规划-最长公共子序列

来源：互联网发布：海关出口数据怎么查编辑：程序博客网时间：2024/05/17 17:43

assuming i,j start from 1;
               0,                                  ( i=0 or j=0)
c(i,j)=   c(i-1,j-1)+1                      ( i,j>0 and A[i]=B[j])
              Max( c[i-1,j),c(i,j-1))      ( i,j>0 and A[i]!=B[j])

时间复杂度O(m*n),空间复杂度O(m*n)

#define MAX 100int C[MAX][MAX];int D[MAX][MAX];//输出子序列void Print_LCSub(char *str,int i,int j){if(i==0|| j==0)return;if(D[i][j]==0){Print_LCSub(str,i-1,j-1);cout<<str[i-1];}else if(D[i][j]==1)Print_LCSub(str,i-1,j);elsePrint_LCSub(str,i,j-1);}void LCSubsequence(char *str1,char *str2){int len1,len2;int i,j;len1=strlen(str1);len2=strlen(str2);for(i=0;i<=len1;i++)C[i][0]=0;for(j=1;j<=len2;j++)C[0][j]=0;for(i=1;i<=len1;i++){for(j=1;j<=len2;j++){if(str1[i-1]==str2[j-1]){C[i][j]=C[i-1][j-1]+1;D[i][j]=0; // left and up}else if(C[i-1][j]>=C[i][j-1]){C[i][j]=C[i-1][j];D[i][j]=1; // up }else{C[i][j]=C[i][j-1];D[i][j]=-1; // left }}}//output matrixfor(i=0;i<=len1;i++){for(j=0;j<=len2;j++)cout<<setw(2)<<C[i][j]<<" ";cout<<endl;}cout<<"MaxLen="<<C[len1][len2]<<endl;Print_LCSub(str1,len1,len2);cout<<endl;}void main(){char str1[200]="xab4c3de89fgh";char str2[200]="ac234egxxh9";//freopen("D:/test.txt","r",stdin);LCSubsequence(str1,str2);}

输出结果：