HDU 1503——Advanced Fruits【LCS最长公共子序列变形 & 合并字符串 & 还原路径】

题目传送门

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Problem Description

The company “21st Century Fruits” has specialized in creating new sorts of fruits by transferring genes from one fruit into the genome of another one. Most times this method doesn’t work, but sometimes, in very rare cases, a new fruit emerges that tastes like a mixture between both of them. A big topic of discussion inside the company is “How should the new creations be called?” A mixture between an apple and a pear could be called an apple-pear, of course, but this doesn’t sound very interesting. The boss finally decides to use the shortest string that contains both names of the original fruits as sub-strings as the new name. For instance, “applear” contains “apple” and “pear” (APPLEar and apPlEAR), and there is no shorter string that has the same property.

A combination of a cranberry and a boysenberry would therefore be called a “boysecranberry” or a “craboysenberry”, for example.

Your job is to write a program that computes such a shortest name for a combination of two given fruits. Your algorithm should be efficient, otherwise it is unlikely that it will execute in the alloted time for long fruit names.

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Input

Each line of the input contains two strings that represent the names of the fruits that should be combined. All names have a maximum length of 100 and only consist of alphabetic characters.

Input is terminated by end of file.

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Output

For each test case, output the shortest name of the resulting fruit on one line. If more than one shortest name is possible, any one is acceptable.

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Sample Input

apple peach ananas banana pear peach

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Sample Output

appleach bananas pearch

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题意：

输入两个长度不超过一百的非空字符串，求一个字符串，使该字符串存在子序列分别是输入的两字符串，要求所求字符串最短。有多个答案时，输出任意一个。

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分析：

根据LCS的原理，将每个字符都进行标记，看两个字符串中对应的字符究竟处于什么状态，然后输出，其标记为公共子串的字符只输出一次即可，也是一道模板题 详细方法见代码。

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AC - code：

#include<iostream> #include<algorithm> #include<cstring> #include<cstdio> #include<sstream> #include<vector> #include<stack> #include<queue> #include<cmath> #include<map> #include<set> using namespace std; typedef long long ll; typedef pair<int,int> pll; const int INF = 0x3f3f3f3f; const int MAXN = 100 + 5; string s1, s2; int dp[MAXN][MAXN]; int dir[MAXN][MAXN]; void Print(int i, int j){ if(i==0 && j==0) return; if(dir[i][j]==2){ Print(i-1, j); cout << s1[i-1]; } else if(dir[i][j]==1){ Print(i-1, j-1); cout << s1[i-1]; } else if(dir[i][j]==3){ Print(i, j-1); cout << s2[j-1]; } } int main(){ ios::sync_with_stdio(false); while(cin >> s1 >> s2){ memset(dp, 0, sizeof(dp)); memset(dir, 0, sizeof(dir)); for(int i = 0; i < s1.length(); i++) dir[i][0]=2; for(int i = 0; i < s2.length(); i++) dir[0][i]=3; for(int i = 1; i <= s1.length(); i++){ for(int j = 1; j <= s2.length(); j++){ if(s1[i-1] == s2[j-1]){ dp[i][j]=dp[i-1][j-1]+1; dir[i][j]=1;//左上方 } else { if(dp[i-1][j]>dp[i][j-1]){ dp[i][j]=dp[i-1][j]; dir[i][j]=2;///来自于左方 } else{ dp[i][j]=dp[i][j-1]; dir[i][j]=3;///来自于上方 } } } } Print(s1.length(), s2.length()); cout << endl; //cout << dp[s1.length()][s2.length()] << endl; } return 0; }