首页 > 代码库 > poj 2264 Advanced Fruits(DP)

poj 2264 Advanced Fruits(DP)

2024-07-10 22:32:26   224人阅读

Advanced Fruits
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 1944 Accepted: 967 Special Judge

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

Source

Ulm Local 1999

题意：

给你两个长度不超过100的字符串A,B。要你找出一个最短的字符串C。使得A,B都是C的子序列(不一定是子串).

思路：

开始看题目里面写的是sub-strings然后看样例百思不得其解。没办法。大胆假设出题人这两个概念分不清。思索片刻后想出了O(n^3)的算法。vis[k][i][j]表示C串长为k时包含了A串的前i-1个字符和B串的前j-1个位置。那么

if(A[i]==B[j])

vis[k+1][i+1][j+1]=1;

else

vis[k+1][i+1][j]=1,vis[k+1][i][j+1]=1;

然后找到最小的k就行了。

比赛的时候各种犯傻。调试了很久才调出来。下来又想了下这道题。觉得先前的做法简直可爱到极致。为何生拉硬扯加个k.直接dp[i][j]表示包含了A串的前i-1个字符和B串的前j-1个位置的C串的最短长度。

if(A[i]==B[j])

dp[i][j]=dp[i-1][j-1]+1;

else

dp[i][j]=min(dp[i][j-1],dp[i-1][j])+1;

然后记录转移的方向就好了。

详细见代码：

#include<iostream>
#include<string.h>
#include<stdio.h>
using namespace std;
const int INF=0x3f3f3f3f;
const int maxn=100010;
int dp[150][150],path[150][150];
char sa[150],sb[150];
void print(int i,int j)
{
    if(i==0&&j==0)
        return;
    if(path[i][j]==1)
    {
        print(i-1,j);
        printf("%c",sa[i]);
    }
    else if(path[i][j]==-1)
    {
        print(i,j-1);
        printf("%c",sb[j]);
    }
    else
    {
        print(i-1,j-1);
        printf("%c",sa[i]);
    }
}
int main()
{
    int i,j,la,lb;

    while(~scanf("%s%s",sa+1,sb+1))
    {
        la=strlen(sa+1);
        lb=strlen(sb+1);
        for(i=1;i<=la;i++)
            dp[i][0]=i,path[i][0]=1;
        for(i=1;i<=lb;i++)
            dp[0][i]=i,path[0][i]=-1;
        for(i=1;i<=la;i++)
            for(j=1;j<=lb;j++)
            {
                dp[i][j]=INF;
                if(sa[i]==sb[j])//相等只用加入一个字符
                    dp[i][j]=dp[i-1][j-1]+1,path[i][j]=0;
                else
                {
                    if(dp[i][j-1]<dp[i-1][j])//增加sb[j]
                        dp[i][j]=dp[i][j-1]+1,path[i][j]=-1;
                    else//增加sa[i]
                        dp[i][j]=dp[i-1][j]+1,path[i][j]=1;
                }
            }
        print(la,lb);
        printf("\n");
    }
    return 0;
}

比赛的做法:

#include <iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
int vis[210][150][150],path[210][150][150];
char sa[150],sb[150];
void print(int st,int i,int j)
{
    //printf("st %d i %d j %d\n",st,i,j);
    if(st==1)
    {
        if(path[st][i][j]==1)
            printf("%c",sa[i-1]);
        else if(path[st][i][j]==-1)
            printf("%c",sb[j-1]);
        else
            printf("%c",sa[i-1]);
            return;
    }
     if(path[st][i][j]==1)
            {
                print(st-1,i-1,j);
                printf("%c",sa[i-1]);
            }
        else if(path[st][i][j]==-1)
           {
               print(st-1,i,j-1);
                printf("%c",sb[j-1]);
           }
           else
           {
               print(st-1,i-1,j-1);
               printf("%c",sa[i-1]);
           }
}
int main()
{
    int la,lb,i,j,k,len;
    while(~scanf("%s%s",sa,sb))
    {
        la=strlen(sa);
        lb=strlen(sb);
        memset(vis,0,sizeof vis);
        len=la+lb;
        vis[1][1][0]=vis[1][0][1]=1,path[1][1][0]=1,path[1][0][1]=-1;
        if (sa[0]==sb[0]) vis[1][1][1]=1,path[1][1][1]=0;
        for(k=1;k<=len;k++)
        {
            if(vis[k][la][lb])
                    break;
                    //printf("k  %d\n",k);
            for(i=0;i<=la;i++)
                for(j=0;j<=lb;j++)
                {
                    if(!vis[k][i][j])
                        continue;
                        //printf("%d %d\n",i,j);
                        if(i<la&&j<lb&&sa[i]==sb[j]&&!vis[k+1][i+1][j+1])
                                vis[k+1][i+1][j+1]=1,path[k+1][i+1][j+1]=0;
                        else
                        {
                            if(!vis[k+1][i+1][j])
                                vis[k+1][i+1][j]=1,path[k+1][i+1][j]=1;
                            if(!vis[k+1][i][j+1])
                                vis[k+1][i][j+1]=1,path[k+1][i][j+1]=-1;

                        }
                }
        }
        //printf("k %d\n",k);
        print(k,la,lb);
        printf("\n");
    }
    return 0;
}


联系
我们