hdu1503 Advanced Fruits

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

最长公共子序列

#include"bits/stdc++.h"
using namespace std;
typedef long long ll;
int dp[1005][1005],flag[1005][1005];
char a[1005],b[1005];
int main(){
    int n;
    while(cin>>a>>b){
        memset(dp,0,sizeof(dp));
        memset(flag,0,sizeof(flag));
        int m = strlen(a);
        int n = strlen(b);
        int m1=m,n1=n;
        int i,j;
        for(i = 1;i <= m;i++)   //第i个之前的最长公共子序列,所以要从1到m，不是0到m-1。
          for(j = 1;j <= n;j++)//第j个之前的最长公共子序列
          {
            if(a[i-1] == b[j-1]){

                dp[i][j] = dp[i-1][j-1] + 1;
                flag[i][j] = 1; //上一步(i-1,j-1),斜向左上
            }
            else if(dp[i-1][j] > dp[i][j-1]){

                dp[i][j] = dp[i-1][j];
                flag[i][j] = 2; //上一步(i-1,j),上方
              }
            else {

                dp[i][j] = dp[i][j-1];
                flag[i][j] = 3;//上一步(i,j-1),左方
              }
          }
        //如何构造最长公共子序列？
        int  k = 0;
        char c[1010];
        memset(c,0,sizeof(c));
        while(m > 0 && n > 0)//从终点按标记的方向反方向往回走
        {
            if(flag[m][n] == 1){ //这个方向来的元素包含在最长公共子序列中
                c[k++] = a[m - 1];
                m--; n--;
            }
            else if(flag[m][n] == 2)
                m--;
            else if(flag[m][n] == 3)
                n--;
        }
        int p1=0,p2=0;
        //cout<
        for(i = k-1;i >= 0;i--){
            while(a[p1]!=c[i]&&p1printf("%c",a[p1]);
                p1++;
            }
            while(b[p2]!=c[i]&&p2printf("%c",b[p2]);
                p2++;
            }
            printf("%c",c[i]);
            p1++,p2++;
        }
        while(a[p1]!=c[i]&&p1printf("%c",a[p1]);
                p1++;
            }
            while(b[p2]!=c[i]&&p2printf("%c",b[p2]);
                p2++;
            }
        printf("\n");

    }
    return 0;
}