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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。
| input | output |
|---|---|
ThesampletextthatcouldbereadedthesameinbothordersArozaupalanalapuazorA |
ArozaupalanalapuazorA |
#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;
}
URAL 1297. Palindrome(后缀数组求最大回文串)
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原文地址：http://blog.csdn.net/xu12110501127/article/details/43063189
