dp--hdu 1503

http://blog.csdn.net/q3498233/article/details/5295082

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. 

解题思路：这道题就是给你两个单词，然后你要把两个单词拼接成一个新单词，使得新单词的子序列中包含两个单词，并且要使这个新单词最短。所以这道题就是求最长公共子序列，并且要记录下子序列的字母，以及他们在主串和副串中的原始位置，之后进行拼接输出。

#include <stdio.h>#include <string.h>#define size 201int Max(int x,int y){return x>y?x:y;}struct rem {int i,j;/*i记录主串位置，j记录副串当前字符位置*/char ch;/*记录当前字符*/};char a[size],b[size];int dp[size][size];rem ans[size];int len;/*指示ans的长度*/void lcs(int m,int n){int i,j;memset(dp,0,sizeof(dp));len = 0;for (i=1;i<=m;i++){for (j=1;j<=n;j++){if(a[i]==b[j])dp[i][j] = dp[i-1][j-1]+1;elsedp[i][j] = Max(dp[i-1][j],dp[i][j-1]);}}if(dp[m][n] == 0)/*如果没有公共序列，直接输出*/printf("%s%s",a,b);else{/*取出最长公共子序列的字母*/i = m,j = n;while (i!=0&&j!=0){if ((dp[i][j] = dp[i-1][j-1]+1)&&a[i] == b[j]){ans[len].i = i;ans[len].j = j;ans[len++].ch = a[i];/*倒序保存最长公共子序列字母*/i--,j--;}else if(dp[i-1][j]>dp[i][j-1])i--;elsej--;}/*取出最长公共子序列的字母*/}}int main(){int a1,b1,i,j,k;while (scanf("%s%s",a+1,b+1)!=EOF){a1 = strlen(a+1);b1 = strlen(b+1);lcs(a1,b1);i = j = 1;for (k=len-1;k>=0;k--){while (i!=ans[k].i){printf("%c",a[i]);i++;}while (j!=ans[k].j){printf("%c",b[j]);j++;}printf("%c",ans[k].ch);i++,j++;}printf("%s%s/n",a+ans[0].i+1,b+ans[0].j+1);}return 0;}