HDU 2594

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http://acm.hdu.edu.cn/showproblem.php?pid=2594

Simpsons’ Hidden Talents

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3323 Accepted Submission(s): 1247

Problem Description

Homer: Marge, I just figured out a way to discover some of the talents we weren’t aware we had.
Marge: Yeah, what is it?
Homer: Take me for example. I want to find out if I have a talent in politics, OK?
Marge: OK.
Homer: So I take some politician’s name, say Clinton, and try to find the length of the longest prefix
in Clinton’s name that is a suffix in my name. That’s how close I am to being a politician like Clinton
Marge: Why on earth choose the longest prefix that is a suffix???
Homer: Well, our talents are deeply hidden within ourselves, Marge.
Marge: So how close are you?
Homer: 0!
Marge: I’m not surprised.
Homer: But you know, you must have some real math talent hidden deep in you.
Marge: How come?
Homer: Riemann and Marjorie gives 3!!!
Marge: Who the heck is Riemann?
Homer: Never mind.
Write a program that, when given strings s1 and s2, finds the longest prefix of s1 that is a suffix of s2.

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.
The lengths of s1 and s2 will be at most 50000.

Sample Input

clintonhomerriemannmarjorie

Sample Output

0rie 3

Source

HDU 2010-05 Programming Contest

题目大意：求一个串的前缀和另一个串的后缀相同的最大长度

有两种解题思路：（1）将两个串拼接成一个串，然后求这个串的浅醉和后缀相等的最大长度。(用next数组的性质)

（2）直接KMP把第一个当成子串，第二个当成主串。匹配，返回的j就是子串的长度。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<vector>
#include<cmath>
#include<stdlib.h>
#include<iomanip>
#include<list>
#include<deque>
#include<map>
#include <stdio.h>
#include <queue>
#include <stack>
#define maxn 10000+5
#define ull unsigned long long
#define ll long long
#define reP(i,n) for(i=1;i<=n;i++)
#define rep(i,n) for(i=0;i<n;i++)
#define cle(a) memset(a,0,sizeof(a))
#define mod 90001
#define PI 3.141592657
#define INF 1<<30
const ull inf = 1LL << 61;
const double eps=1e-5;

using namespace std;

bool cmp(int a,int b){
    return a>b;
}
int f[100005];
char p[100005],t[50005];

void getFail(char* p, int* f)
{
int m = strlen(p);
f[0] = 0;
f[1] = 0;
for (int i = 1; i < m; i++)
{
int j = f[i];
while (j && p[i] != p[j])
{
j = f[j];
}
f[i + 1] = p[i] == p[j] ? j + 1 : 0;
}
}
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    while(~scanf("%s%s",p,t))
{
int len1=strlen(p);
int len2=strlen(t);
int len=len1+len2;
strcat(p,t);///字符串拼接
getFail(p,f);
int k=0;
while(f[len]>len1||f[len]>len2)
{
len=f[len];
}
len=f[len];
if(len==0)printf("%d\n",0);
else
{
for(int i=0;i<len;i++)
printf("%c",p[i]);
printf(" %d\n",len);
}
}
    return 0;
}

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<vector>
#include<cmath>
#include<stdlib.h>
#include<iomanip>
#include<list>
#include<deque>
#include<map>
#include <stdio.h>
#include <queue>

#define maxn 1000010
#define ull unsigned long long
#define ll long long
#define reP(i,n) for(i=1;i<=n;i++)
#define rep(i,n) for(i=0;i<n;i++)
#define cle(a) memset(a,0,sizeof(a))
#define mod 90001
#define PI 3.141592657
#define INF 1<<30
const ull inf = 1LL << 61;
const double eps=1e-5;

using namespace std;

bool cmp(int a,int b){
    return a>b;
}
char p[50005],t[50005];
int f[50005];
void getFail(char *p,int *f,int m)
{
f[0]=0;
f[1]=0;
for(int i=1;i<m;i++)
{
int j=f[i];
while(j&&p[i]!=p[j])
{
j=f[j];
}
f[i+1]=p[i]==p[j]?j+1:0;
}
}
int find(char *t,char *p,int *f)
{
int n=strlen(t),m=strlen(p);
getFail(p,f,m);
int j=0;
for(int i=0;i<n;i++)
{
while(j&&p[j]!=t[i])
{
j=f[j];
}
if(p[j]==t[i])j++;
}
return j;
}
int main()
{

while(~scanf("%s%s",p,t))
{
cle(f);
     int k=find(t,p,f);
     if(k==0)
printf("%d\n",0);
 else
{
for(int i=0;i<k;i++)
printf("%c",p[i]);
printf(" %d\n",k);
}
cle(p),cle(t);
}
    return 0;
}

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