两个题目意思差不多，都是让求最长公共子串，只不过poj那个让输出长度，而URAL那个让输出一个任意的最长的子串。
解体思路：
Long Long Message
| Time Limit: 4000MS | Memory Limit: 131072K | |
| Total Submissions: 22313 | Accepted: 9145 | |
| Case Time Limit: 1000MS | ||
Description
The little cat is majoring in physics in the capital of Byterland. A piece of sad news comes to him these days: his mother is getting ill. Being worried about spending so much on railway tickets (Byterland is such a big country, and he has to spend 16 shours
on train to his hometown), he decided only to send SMS with his mother.
The little cat lives in an unrich family, so he frequently comes to the mobile service center, to check how much money he has spent on SMS. Yesterday, the computer of service center was broken, and printed two very long messages. The brilliant little cat soon
found out:
1. All characters in messages are lowercase Latin letters, without punctuations and spaces.
2. All SMS has been appended to each other – (i+1)-th SMS comes directly after the i-th one – that is why those two messages are quite long.
3. His own SMS has been appended together, but possibly a great many redundancy characters appear leftwards and rightwards due to the broken computer.
E.g: if his SMS is “motheriloveyou”, either long message printed by that machine, would possibly be one of “hahamotheriloveyou”, “motheriloveyoureally”, “motheriloveyouornot”, “bbbmotheriloveyouaaa”, etc.
4. For these broken issues, the little cat has printed his original text twice (so there appears two very long messages). Even though the original text remains the same in two printed messages, the redundancy characters on both sides would be possibly different.
You are given those two very long messages, and you have to output the length of the longest possible original text written by the little cat.
Background:
The SMS in Byterland mobile service are charging in dollars-per-byte. That is why the little cat is worrying about how long could the longest original text be.
Why ask you to write a program? There are four resions:
1. The little cat is so busy these days with physics lessons;
2. The little cat wants to keep what he said to his mother seceret;
3. POJ is such a great Online Judge;
4. The little cat wants to earn some money from POJ, and try to persuade his mother to see the doctor :(
Input
Two strings with lowercase letters on two of the input lines individually. Number of characters in each one will never exceed 100000.
Output
A single line with a single integer number – what is the maximum length of the original text written by the little cat.
Sample Input
yeshowmuchiloveyoumydearmotherreallyicannotbelieveit yeaphowmuchiloveyoumydearmother
Sample Output
27
#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <ctime>
#include <map>
#include <set>
#define eps 1e-9
///#define M 1000100
///#define LL __int64
#define LL long long
///#define INF 0x7ffffff
#define INF 0x3f3f3f3f
#define PI 3.1415926535898
#define zero(x) ((fabs(x)<eps)?0:x)
#define mod 1000000007
#define Read() freopen("autocomplete.in","r",stdin)
#define Write() freopen("autocomplete.out","w",stdout)
#define Cin() ios::sync_with_stdio(false)
using namespace std;
inline int read()
{
char ch;
bool flag = false;
int a = 0;
while(!((((ch = getchar()) >= '0') && (ch <= '9')) || (ch == '-')));
if(ch != '-')
{
a *= 10;
a += ch - '0';
}
else
{
flag = true;
}
while(((ch = getchar()) >= '0') && (ch <= '9'))
{
a *= 10;
a += ch - '0';
}
if(flag)
{
a = -a;
}
return a;
}
void write(int a)
{
if(a < 0)
{
putchar('-');
a = -a;
}
if(a >= 10)
{
write(a / 10);
}
putchar(a % 10 + '0');
}
const int maxn = 200050;
int wa[maxn], wb[maxn], wv[maxn], ws1[maxn];
int sa[maxn];
int cmp(int *r, int a, int b, int l)
{
return r[a] == r[b] && r[a+l] == r[b+l];
}
void da(int *r, int *sa, int n, int m)
{
int i, j, p, *x = wa, *y = wb;
for(i = 0; i < m; i++) ws1[i] = 0;
for(i = 0; i < n; i++) ws1[x[i] = r[i]]++;
for(i = 1; i < m; i++) ws1[i] += ws1[i-1];
for(i = n-1; i >= 0; i--) sa[--ws1[x[i]]] = i;
for(j = 1, p = 1; p < n; j <<= 1, m = p)
{
for(p = 0, i = n-j; i < n; i++) y[p++] = i;
for(i = 0; i < n; i++)
if(sa[i] >= j) y[p++] = sa[i]-j;
for(i = 0; i < n; i++) wv[i] = x[y[i]];
for(i = 0; i < m; i++) ws1[i] = 0;
for(i = 0; i < n; i++) ws1[wv[i]]++;
for(i = 1; i < m; i++) ws1[i] += ws1[i-1];
for(i = n-1; i >= 0; i--) sa[--ws1[wv[i]]] = y[i];
for(swap(x, y), p = 1, x[sa[0]] = 0, i = 1; i < n; i++)
x[sa[i]] = cmp(y, sa[i-1], sa[i], j)?p-1:p++;
}
}
int rank[maxn], height[maxn];
void calheight(int *r, int *sa, int n)
{
int i, j, k = 0;
for(i = 1; i <= n; i++) rank[sa[i]] = i;
for(int i = 0; i < n; height[rank[i++]] = k)
for(k?k--:0, j = sa[rank[i]-1]; r[i+k] == r[j+k]; k++);
return ;
}
int dp[maxn][30];
void RMQ(int len)
{
for(int i = 1; i <= len; i++)
dp[i][0] = height[i];
for(int j = 1; 1<<j <= maxn; j++)
{
for(int i = 1; i+(1<<j)-1 <= len; i++)
dp[i][j] = min(dp[i][j-1], dp[i+(1<<(j-1))][j-1]);
}
}
int lg[maxn];
int querry(int l, int r)
{
int k = lg[r-l+1];
return min(dp[l][k], dp[r-(1<<k)+1][k]);
}
void init()
{
lg[0] = -1;
for (int i = 1; i < maxn; ++i)
lg[i] = lg[i>>1] + 1;
}
int seq[2*maxn];
char str1[maxn], str2[maxn];
void Del(int n, int len1, int len2)
{
int xp = 0;
for(int i = 2; i <= n; i++)
{
if(xp < height[i])
{
int fx = sa[i-1];
int fy = sa[i];
int xx = max(fx, fy);
int yy = min(fx, fy);
if(xx > len1 && yy < len1) xp = height[i];
}
}
printf("%d\n", xp);
}
int main()
{
///init();
while(~scanf("%s %s",str1, str2))
{
int len1 = strlen(str1);
int len2 = strlen(str2);
int n = 0;
for(int i = 0; i < len1; i++) seq[n++] = str1[i];
seq[n++] = 1;
for(int i = 0; i < len2; i++) seq[n++] = str2[i];
seq[n] = 0;
da(seq, sa, n+1, 130);
calheight(seq, sa, n);
Del(n, len1, len2);
}
return 0;
}