Description
Input
Output
Sample Input
| input | output |
|---|---|
|
ThesampletextthatcouldbereadedthesameinbothordersArozaupalanalapuazorA
|
ArozaupalanalapuazorA
|
题意：求最长回文子串。
思路：穷举每一位，然后计算以这个字符为中心的最长回文子串。注意这里要分两种情况，一是回文子串的长度为奇数，二是长度为偶数。两种情况都可以转化为求一个后缀和一个反过来写的后缀的最长公共前缀。具体的做法是：将整个字符串反过来写在原字符串后面，中间用一个特殊的字符隔开。这样就把问题变为求这个新的字符串的某两个后缀的最长公共前缀。然后在查找最长公共前缀lcp的时候利用到了RMQ，我们知道对于两个后缀j和k，设rank[j]<rank[k]，则不难证明后缀j和k的LCP长度等于height[rank[j]+1]，height[rank[j]+2]，...,height[rank[k]]中的最小值，
即RMQ(height, rank[j]+1, rank[k])，这是通过height[]数组的两两比较做到的
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
const int maxn = 2010;
int sa[maxn]; //SA数组，表示将S的n个后缀从小到大排序后把排好序的
//的后缀的开头位置顺次放入SA中
int t1[maxn], t2[maxn], c[maxn];
int rank[maxn], height[maxn];
int s[maxn], r[maxn];
char str[maxn];
int st[maxn][20];
void build_sa(int s[], int n, int m) {
int i, j, p, *x = t1, *y = t2;
for (i = 0; i < m; i++) c[i] = 0;
for (i = 0; i < n; i++) c[x[i] = s[i]]++;
for (i = 1; i < m; i++) c[i] += c[i-1];
for (i = n-1; i >= 0; i--) sa[--c[x[i]]] = i;
for (j = 1; j <= n; j <<= 1) {
p = 0;
for (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 < m; i++) c[i] = 0;
for (i = 0; i < n; i++) c[x[y[i]]]++;
for (i = 1; i < m; i++) c[i] += c[i-1];
for (i = n-1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i];
swap(x, y);
p = 1, x[sa[0]] = 0;
for (i = 1; i < n; i++)
x[sa[i]] = y[sa[i-1]] == y[sa[i]] && y[sa[i-1]+j] == y[sa[i]+j] ? p-1 : p++;
if (p >= n) break;
m = p;
}
}
void getHeight(int s[],int n) {
int i, j, k = 0;
for (i = 0; i <= n; i++)
rank[sa[i]] = i;
for (i = 0; i < n; i++) {
if (k) k--;
j = sa[rank[i]-1];
while (s[i+k] == s[j+k]) k++;
height[rank[i]] = k;
}
}
void ST(int n) {
for (int i = 1; i <= n; i++)
st[i][0] = i;
for (int j = 1; (1<<j) <= n; j++) {
for (int i = 1; i + (1<<j) <= n; i++) {
int p = st[i][j-1];
int q = st[i+(1<<(j-1))][j-1];
st[i][j] = height[p] > height[q] ? q : p;
}
}
}
int RMQ(int i, int j) {
int k = 0;
if (i > j)
swap(i, j);
i++;
while ((1<<(k+1)) <= j-i+1) k++;
i = st[i][k];
j = st[j-(1<<k)+1][k];
return min(height[i], height[j]);
}
int main() {
while (scanf("%s", str) != EOF) {
int len = strlen(str);
int n = 2 * len + 1;
for (int i = 0; i < len; i++)
r[i] = str[i];
r[len] = 1;
for (int i = 0; i < len; i++)
r[i+len+1] = str[len-1-i];
r[n] = 0; //notice
build_sa(r, n+1, 128);
getHeight(r, n);
ST(n);
int mid = n >> 1;
int ans = 0, cur = 0;
for (int i = 0; i < mid; i++) {
int j = RMQ(rank[i], rank[n-i-1]); //奇对称
if ((j<<1) - 1 > ans) {
ans = (j<<1) - 1;
cur = i - j + 1;
}
if (i) {
j = RMQ(rank[i], rank[n-i]); //偶对称
if ((j << 1) > ans) {
ans = j << 1;
cur = i - j;
}
}
}
for (int i = cur; i < cur + ans; i++)
printf("%c", r[i]);
printf("
");
}
return 0;
}
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