题解报告：hdu 1503 Advanced Fruits（LCS加强版）

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. 

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. 

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.

Sample Input

apple peach

ananas banana

pear peach

Sample Output

appleach

bananas

pearch

解题思路：题目的意思就是将给定的两个字符串合并成一个最短的新串，即要求它们的最长公共子序列在新串中只出现1次，并且新串的子序列中包含原来两个字符串。LCS的灵活应用。做法：假设第一个字符串为s1，第二个字符串为s2，在计算LCS同时，用一个二维数组标记当前i、j字符的相对状态：mp[i][j]=-1表示s1前i个字符中包含公共子序列的长度L1不小于s2前j个字符中的L2；mp[i][j]=0表示s1前i个字符中包含公共子序列的长度L1与s2前j个字符中的L2相等；mp[i][j]=1表示s1前i个字符包含公共子序列的长度L1小于s2前j个字符中的L2。接下来就可以利用递归的特性，通过对标记的判断，从后往前递归，回溯时就会根据标记依次输出其正确位置。解释一下递归部分的含义：如果当前mp[i][j]为-1，说明s1前i个字符中包含公共子序列的长度不小于s2前j个字符，并且此时s1[i]!=s2[j]，即i>j，所以先输出s1[i]，然后再在s1前i-1个字符中往前找，s2保持j位置不变（实际输出是在回溯时从前往后输出的，这样就比较好理解了），其他两种情况与此相同。注意：要初始化mp[i∈[1～len1]][0]=-1，mp[0][j∈[1,len2]]=1，含义与上面相同，为什么呢？举个栗子：假设递归到i=3,j=0时，由于一开始就将mp数组全部初始化为0，即mp[3][0]=0，这个时候满足第9行的条件，然后继续i-1=2,j-1=-1，此时已越界并且陷入死循环中，事实上初始为-1后即mp[3][0]=-1满足第8行这个条件，继续i-1=2，j=0不变递归下去就不会出错了，当i和j都为0时就可以开始回溯即return依次弹出栈顶信息并输出对应位置的字符。此题还有一个特点就是special judge特判表示答案不唯一，譬如样例三，新串还可以为peachr，其子序列同样包含原来两个字符串。

AC代码：

#include<bits/stdc++.h>
using namespace std;
const int maxn=105;
char s1[maxn],s2[maxn];
int len1,len2,dp[maxn][maxn],mp[maxn][maxn];
void print_LCS(int i,int j){
    if(!i&&!j)return;//如果i，j都为0，则直接返回
    else if(mp[i][j]==-1){print_LCS(i-1,j);putchar(s1[i]);}
    else if(mp[i][j]==0){print_LCS(i-1,j-1);putchar(s1[i]);}
    else {print_LCS(i,j-1);putchar(s2[j]);}
}
int main(){
    while(~scanf("%s%s",s1+1,s2+1)){
        memset(dp,0,sizeof(dp));
        memset(mp,0,sizeof(mp));//初始状态默认两个字符串是相等的
        len1=strlen(s1+1),len2=strlen(s2+1);
        for(int i=1;i<=len1;++i)mp[i][0]=-1;
        for(int j=1;j<=len2;++j)mp[0][j]=1;
        for(int i=1;i<=len1;++i){
            for(int j=1;j<=len2;++j){
                if(s1[i]==s2[j])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],mp[i][j]=-1;
                else dp[i][j]=dp[i][j-1],mp[i][j]=1;
            }
        }
        print_LCS(len1,len2);
        puts("");
    }
    return 0;
}