Description
Input
Output
Sample Input
| input | output |
|---|---|
|
ThesampletextthatcouldbereadedthesameinbothordersArozaupalanalapuazorA
|
ArozaupalanalapuazorA
|
題意：求最長回文子串。
思路：窮舉每一位，然后計算以這個字符為中心的最長回文子串。注意這里要分兩種情況，一是回文子串的長度為奇數，二是長度為偶數。兩種情況都可以轉化為求一個后綴和一個反過來寫的后綴的最長公共前綴。具體的做法是：將整個字符串反過來寫在原字符串后面，中間用一個特殊的字符隔開。這樣就把問題變為求這個新的字符串的某兩個后綴的最長公共前綴。然后在查找最長公共前綴lcp的時候利用到了RMQ，我們知道對于兩個后綴j和k，設rank[j]<rank[k]，則不難證明后綴j和k的LCP長度等于height[rank[j]+1]，height[rank[j]+2]，...,height[rank[k]]中的最小值，
即RMQ(height, rank[j]+1, rank[k])，這是通過height[]數組的兩兩比較做到的
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
const int maxn = 2010;
int sa[maxn]; //SA數組，表示將S的n個后綴從小到大排序后把排好序的
//的后綴的開頭位置順次放入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]); //奇對稱
if ((j<<1) - 1 > ans) {
ans = (j<<1) - 1;
cur = i - j + 1;
}
if (i) {
j = RMQ(rank[i], rank[n-i]); //偶對稱
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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