Advanced Fruits(好题,LCS的模拟)
Advanced Fruits
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2052 Accepted Submission(s): 1053
Special Judge
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 is terminated by end of file.
#include<stdio.h>
#include<string.h>
#define MAX(x,y) x>y?x:y
const int MAXN=;
int dp[MAXN][MAXN];
char s1[MAXN],s2[MAXN],ans[MAXN*];
int t1,t2,t;
void LCS(){
memset(dp,,sizeof(dp));
t1=strlen(s1+);t2=strlen(s2+);
for(int i=;i<=t1;i++){
for(int j=;j<=t2;j++){
if(s1[i]==s2[j])dp[i][j]=dp[i-][j-]+;
else{
dp[i][j]=MAX(dp[i-][j],dp[i][j-]);
}
}
}
}
void add(char a){
ans[t++]=a;
ans[t]='\0';
}
void print(){
while(dp[t1][t2]){
if(s1[t1]==s2[t2]){
add(s1[t1]);
t1--;t2--;
}
else{
if(dp[t1-][t2]>dp[t1][t2-]){
add(s1[t1]);
t1--;
}
else{
add(s2[t2]);
t2--;
}
}
}
while(t1>)add(s1[t1--]);
while(t2>)add(s2[t2--]);
}
int main(){
while(~scanf("%s%s",s1+,s2+)){
t=;
LCS();
print();
for(int i=t-;i>=;i--)printf("%c",ans[i]);
puts("");
}
return ;
}
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