HDOJ-1503 Advanced Fruits

来源：互联网发布：中国出口贸易数据分析编辑：程序博客网时间：2024/05/17 23:33

这道题是最长公共子序列的变形。
用dp(i)(j)表示s1的前i个字符和s2的前j个字符，构成的shortest name .
状态转移方程:1.若s1(i) == s2(j)，那么dp(i)(j) = dp(i-1)(j-1) + 1;
2.若s1(i) != s2(j), 那么dp(i)(j) = min(dp(i-1)(j), dp(i)(j-1)) + 1;

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>using namespace std;int dp[105][105];char s1[105], s2[105];void print(int n1, int n2, string &ans){      if(n1 == 0 && n2 == 0)         return ;      if(s1[n1] == s2[n2])      {          ans += s1[n1];          print(n1-1, n2-1, ans);      }      else      if(n1 >= 1 && dp[n1][n2] == dp[n1-1][n2] + 1)      {          ans += s1[n1];          print(n1-1, n2, ans);      }      else      if(n2 >= 1 && dp[n1][n2] == dp[n1][n2-1] + 1)      {          ans += s2[n2];          print(n1, n2-1, ans);      }}int main(){   // freopen("in.txt", "r", stdin);    while(scanf("%s %s", s1+1, s2+1) == 2)    {        string ans;        int n1 = strlen(s1+1), n2 = strlen(s2+1);        for(int i = 1; i <= n2; i++)            dp[0][i] = i;        for(int i = 1; i <= n1; i++)            dp[i][0] = i;        for(int i = 1; i <= n1; i++)            for(int j = 1; j <= n2; j++)        {            if(s1[i] == s2[j])                dp[i][j] = dp[i-1][j-1] + 1;            else                dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + 1;        }        print(n1, n2, ans);        for(int i = ans.size()-1; i >= 0; i--)            cout << ans[i];        cout << endl;    }    return 0;}

0 0