dp-LCS(递归输出最短合串)

最新推荐文章于 2024-07-02 19:42:08 发布

转载最新推荐文章于 2024-07-02 19:42:08 发布 · 112 阅读

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原文链接：http://www.cnblogs.com/ccut-ry/p/7476216.html

本文介绍了一种用于合并两种水果名称的算法，旨在找到最短的字符串，该字符串同时包含两种水果名称作为子串。通过类似LCS的方法实现，并提供了一个具体的代码示例。

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

题目大意 ：

　　找到一个最小的，同时包含两个子串的串，并输出这个最小的子串，这个题目的思想的就是类似于 LCS ， 但唯一要有区别的地方，就是在通过比较字母时， 同时对他们进行标记 ，最后输出的时候递归输出 。

代码示例：

/*
 * Author:  ry 
 * Created Time:  2017/9/4 21:04:24
 * File Name: 1.cpp
 */
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <time.h>
using namespace std;
const int mm = 1e6+5;
#define Max(a,b) a>b?a:b
#define Min(a,b) a>b?b:a
#define ll long long

ll t_cnt;
void t_st(){t_cnt=clock();}
void t_ot(){printf("you spent : %lldms\n", clock()-t_cnt);}
//开始t_st();
//结束t_ot();

char a[105], b[105];
int dp[105][105];
int mark[105][105];

void fun(int i , int j){
    if (!i && !j) return;
    
    if (mark[i][j] == 0){
        fun(i-1,j-1);
        printf("%c", a[i]);
    }
    else if (mark[i][j] == 1){
        fun(i-1, j);
        printf("%c", a[i]);
    }
    else {
        fun(i, j-1);
        printf("%c", b[j]);
    }
    
}

int main() {
    
    while (~scanf("%s%s", a, b)){
        int len1 = strlen(a);
        int len2 = strlen(b);
        for(int i = len1; i > 0; i--)
            a[i] = a[i-1];
        for(int i = len2; i > 0; i--)
            b[i] = b[i-1];
        
        memset(dp, 0, sizeof(dp));
        for(int i = 1; i <= len1; i++)
            mark[i][0] = 1;
        for(int j = 1; j <= len2; j++)
            mark[0][j] = -1;
        
        for (int i = 1; i <= len1; i++){
            for(int j = 1; j <= len2; j++) {
                if (a[i] == b[j]) {
                    dp[i][j] = dp[i-1][j-1] + 1;
                    mark[i][j] = 0;
                }        
                else if (dp[i-1][j] >= dp[i][j-1]){
                    dp[i][j] = dp[i-1][j];
                    mark[i][j] = 1;
                }
                else {
                    dp[i][j] = dp[i][j-1];
                    mark[i][j] = -1;
                }
            }
        }
        
        int i = len1, j = len2;
        fun(i, j);
        printf("\n")
    }
    return 0;
}

 
 

转载于:https://www.cnblogs.com/ccut-ry/p/7476216.html