HDU 2594 Simpsons’ Hidden Talents （字串-KMP 字首與字尾）
Simpsons’ Hidden TalentsTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 15010 Accepted Submission(s): 5148 Problem Description Homer: Marge, I just figured out a way to discover some of the talents we weren’t aware we had.
Input Input consists of two lines. The first line contains s1 and the second line contains s2. You may assume all letters are in lowercase.
Output Output consists of a single line that contains the longest string that is a prefix of s1 and a suffix of s2, followed by the length of that prefix. If the longest such string is the empty string, then the output should be 0.
Sample Input clinton homer riemann marjorie
Sample Output 0 rie 3
Source HDU 2010-05 Programming Contest
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哈哈哈哈，終於自己做出了演算法題了。
直接把這兩個字串合併起來就行，但有一點要注意就是比如
abc abcabc
abcabc 6
所以應該限制一下長度，不能超過原本字串的長度
但是，發現了一組資料不對，但其他做法也不對，
abc dabcd 這個應該輸出0 結果是輸出abc 3
後面程式碼可以解決此錯誤。
資料水啊
程式碼實現：能AC但最後一個樣例沒過
#include<cstdio>
#include<iostream>
#include<fstream>
#include<algorithm>
#include<functional>
#include<cstring>
#include<string>
#include<cstdlib>
#include<iomanip>
#include<numeric>
#include<cctype>
#include<cmath>
#include<ctime>
#include<queue>
#include<stack>
#include<list>
#include<set>
#include<map>
using namespace std;
const int N = 1000002;
int nxt[N];
string S,T,T1;
int tlen;
void getNext()
{
int j, k;
j = 0; k = -1;
nxt[0] = -1;
while(j < tlen)
if(k == -1 || T[j] == T[k])
{
nxt[++j] = ++k;
if (T[j] != T[k])
nxt[j] = k;
}
else
k = nxt[k];
}
int main()
{
while(cin>>T1>>S)
{
T=T1+S;
tlen= T.length();
getNext();
if(nxt[tlen]==0)
{
printf("0\n") ;
continue;
}
int ans=min(T1.length(),S.length());
if(nxt[tlen]<=T1.length()&&nxt[tlen]<=S.length())
ans=nxt[tlen];
cout<<T.substr(0,ans)<<" "<<ans<<endl;
}
return 0;
}
可以過最後一個樣例
#include<cstdio>
#include<iostream>
#include<fstream>
#include<algorithm>
#include<functional>
#include<cstring>
#include<string>
#include<cstdlib>
#include<iomanip>
#include<numeric>
#include<cctype>
#include<cmath>
#include<ctime>
#include<queue>
#include<stack>
#include<list>
#include<set>
#include<map>
using namespace std;
const int N = 1000002;
int nxt[N];
string S,T,T1;
int tlen;
void getNext()
{
int j, k;
j = 0; k = -1;
nxt[0] = -1;
while(j < tlen)
if(k == -1 || T[j] == T[k])
{
nxt[++j] = ++k;
if (T[j] != T[k])
nxt[j] = k;
}
else
k = nxt[k];
}
int main()
{
while(cin>>T1>>S)
{
T=T1+S;
tlen= T.length();
getNext();
int minn=min(T1.length(),S.length());
int index=nxt[tlen];
while (index>minn)//若超出長度
index=nxt[index];//向前迭代尋找
if(index==0)
{
printf("0\n") ;
continue;
}
cout<<T.substr(0,index)<<" "<<index<<endl;
}
return 0;
}