题目
字符串的LCS,输出解我比较不会,dp的时候记录从哪里转移来的,之后要一步一步转移回去把解存起来然后输出。
#include<cstdio> #include<cstring> #define N 102 #define M 32 int n,m,dp[N][N],ans[N][N]; char s1[N][M],s2[N][M],str[N][M]; void solve(){ memset(dp,0,sizeof dp); memset(ans,0,sizeof ans); int i,j,t,ti; for (i=1;i<=n;i++) for (j=1;j<=m;j++) if (strcmp(s1[i-1],s2[j-1])==0) dp[i][j]=dp[i-1][j-1]+1; else if(dp[i-1][j]>dp[i][j-1]){ dp[i][j]=dp[i-1][j]; ans[i][j]=1;//从s1转移来 }else dp[i][j]=dp[i][j-1]; i=n,j=m; t=dp[n][m]; while(i&&j) if(strcmp(s1[i-1],s2[j-1])==0){ memcpy(str[--t],s1[i-1],sizeof s1[i-1]); i--;j--; } else{ ti=i; i-=ans[i][j]; j-=1^ans[ti][j];//ans=0则是s2转移来的,故j-- } for(i=0;i<dp[n][m];i++) printf("%s ",str[i]); printf(" "); } int main(){ while (~scanf("%s",&s1[n])) if (s1[n][0]=='#'){ while (scanf("%s",&s2[m])&&s2[m][0]!='#') m++; solve(); n=m=0; } else n++; return 0; }