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经典的动态规划题目,动态转移方程比较简单。动态转移方程为: f[i][j]=max{f[i-1,j],f[i,j-1],f[i-1,j-1]+1|s1[i]==s2[j]} f[0][0]=0,f[0][i]=0,f[i][0]=0 程序仅供参考: #include <iostream> #include <string> using namespace std; const int maxn=500; int max(int x,int y) { return x>y?x:y; } int main() { string s1,s2; while(cin>>s1>>s2){ int f[maxn][maxn]={0},len1=s1.length(),len2=s2.length(); for(int i=1;i<=len1;i++) for(int j=1;j<=len2;j++){ f[i][j]=max(f[i-1][j],f[i][j-1]); if(s1[i-1]==s2[j-1]) f[i][j]=max(f[i][j],f[i-1][j-1]+1); } cout<<f[len1][len2]<<endl; } return 0; } Followed by:
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