题目大意：给你一串字符串，让你求出来它存在的最长连续的回文串。
解题思路：先把字符串逆序加到数组中，然后用后缀数组求解。两种方法：1，枚举排名，直接比较rank相同的字符串的位置差是不是len。如果是的话，就记录求解；2，枚举地址，求第i地址与第2*len-i+1的lcp的最大值。
PS：需要注意如果多解输出靠前的字符串。
两种写法写在了一起，分别是Del，和Del1函数。
1297. Palindrome
Time limit: 1.0 second
Memory limit: 64 MB
The “U.S. Robots” HQ has just received a rather alarming anonymous letter. It states that the agent from the competing ?Robots Unlimited? has infiltrated into “U.S. Robotics”. ?U.S. Robots? security
service would have already started an undercover operation to establish the agent’s identity, but, fortunately, the letter describes communication channel the agent uses. He will publish articles containing stolen data to the “Solaris” almanac. Obviously,
he will obfuscate the data, so “Robots Unlimited” will have to use a special descrambler (“Robots Unlimited” part number NPRx8086, specifications are kept secret).
Having read the letter, the “U.S. Robots” president recalled having hired the “Robots Unlimited” ex-employee John Pupkin. President knows he can trust John, because John is still angry at being mistreated
by “Robots Unlimited”. Unfortunately, he was fired just before his team has finished work on the NPRx8086 design.
So, the president has assigned the task of agent’s message interception to John. At first, John felt rather embarrassed, because revealing the hidden message isn’t any easier than finding a needle in
a haystack. However, after he struggled the problem for a while, he remembered that the design of NPRx8086 was still incomplete. “Robots Unlimited” fired John when he was working on a specific module, the text direction detector. Nobody else could finish that
module, so the descrambler will choose the text scanning direction at random. To ensure the correct descrambling of the message by NPRx8086, agent must encode the information in such a way that the resulting secret message reads the same both forwards and
backwards.
In addition, it is reasonable to assume that the agent will be sending a very long message, so John has simply to find the longest message satisfying the mentioned property.
Your task is to help John Pupkin by writing a program to find the secret message in the text of a given article. As NPRx8086 ignores white spaces and punctuation marks, John will remove them from the
text before feeding it into the program.
Input
The input consists of a single line, which contains a string of Latin alphabet letters (no other characters will appear in the string). String length will not exceed 1000 characters.
Output
The longest substring with mentioned property. If there are several such strings you should output the first of them.
Sample
| input | output |
|---|---|
ThesampletextthatcouldbereadedthesameinbothordersArozaupalanalapuazorA |
ArozaupalanalapuazorA |
Problem Author: Eugene Krokhalev
Problem Source: IX Open Collegiate Programming Contest of the High School Pupils (13.03.2004)
Tags: (
#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 = 20050;
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;
}
char str[maxn];
int seq[maxn];
void Del1(int n)
{
int Max = 0;
int pos = 0;
for(int i = 0; i < n; i++)
{
int l = rank[i];
int r = rank[2*n-i+1];
if(l > r) swap(l, r);
int tmp = querry(l+1, r);
if(tmp*2 > Max)
{
Max = tmp*2;
pos = i;
}
l = rank[i];
r = rank[2*n-i];
if(l > r) swap(l, r);
tmp = querry(l+1, r);
if(tmp*2-1 > Max)
{
Max = tmp*2-1;
pos = i;
}
}
for (int i = pos-Max/2; Max--; ++i)
putchar(seq[i]);
puts("");
return ;
}
void Del(int n, int len)
{
int Max = 1;
int pos = 0;
for(int i = 1; i <= n; i++)
{
int tmp = height[i];
int l = sa[i-1];
int r = sa[i];
if(l > r) swap(l, r);
if(l >= len || r <= len) continue;
if(l+tmp != n-r) continue;
if(tmp > Max)
{
Max = tmp;
pos = l;
}
else if(tmp == Max && l < pos)
pos = l;
}
for(int i = pos, ans = 0; ans < Max; i++, ans++)
printf("%c",seq[i]);
puts("");
return ;
}
int main()
{
///init();
while(~scanf("%s", str))
{
int len = strlen(str);
int ans = 0;
for(int i = 0; i < len; i++) seq[ans++] = str[i];
seq[ans++] = 1;
for(int i = len-1; i >= 0; i--) seq[ans++] = str[i];
seq[ans] = 0;
int Len = ans;
da(seq, sa, Len+1, 128);
calheight(seq, sa, Len);
Del(ans, len);///枚举位置，判断lcp是否为位置差;
///Del1(len);///枚举位置，判断区间的lcp;
}
return 0;
}