codefores-335B-求最长的回文序列

B. Palindrome

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Given a string s, determine if it contains any palindrome of length exactly 100 as a subsequence. If it has any, print any one of them. If it doesn't have any, print a palindrome that is a subsequence of s and is as long as possible.

Input

The only line of the input contains one string s of length n (1 ≤ n ≤ 5·104) containing only lowercase English letters.

Output

If s contains a palindrome of length exactly 100 as a subsequence, print any palindrome of length 100 which is a subsequence of s. If sdoesn't contain any palindromes of length exactly 100, print a palindrome that is a subsequence of s and is as long as possible.

If there exists multiple answers, you are allowed to print any of them.

Sample test(s)

input

bbbabcbbb

output

bbbcbbb

input

rquwmzexectvnbanemsmdufrg

output

rumenanemur

代码：

#include <iostream>#include <string>#include <stdio.h>#include <algorithm>using namespace std;#define MAXN 111111char s[MAXN];int N = 0;#define MAXM 111int dp[MAXN][111], prefi[MAXN][111], prefj[MAXN][111];int last[MAXN][MAXM];int best = 0;char tmp[MAXN];char ans[MAXN];int main() {scanf("%s", s);N = strlen(s);for (int i = 0; i < 26; i ++)last[0][i] = -1;for (int i = 0; i < N; i ++) {for (int j = 0; j < 26; j ++) {last[i + 1][j] = last[i][j];last[i + 1][s[i] - 'a'] = i;}}for (int j = 0; j < MAXN; j ++) {dp[0][j] = -1;}dp[0][0] = N;for (int i = 0; i < N; i ++)for (int j = 0; j < 55; j ++)if (dp[i][j] >= i) {int len = 2 * j + (dp[i][j] > i);if(len <= 100 && best < len) {best = len;int ii = i, jj = j;int n = 0;while(jj) {int ni = prefi[ii][jj], nj = prefj[ii][jj];if (nj != jj) tmp[n++] = s[ni];ii = ni; jj = nj;}int A = 0;for (int k = 0; k < n; k ++)ans[A++] = tmp[n - 1 - k];if (dp[i][j] > i) ans[A++] = s[i];for (int k = 0; k < n; k ++)ans[A++] = tmp[k];}if(dp[i][j] > dp[i+1][j]) {prefi[i+1][j] = i; prefj[i+1][j] = j;dp[i+1][j] = dp[i][j];}if(last[dp[i][j]][s[i]-'a'] > dp[i+1][j+1]) {prefi[i+1][j+1] = i; prefj[i+1][j+1] = j;dp[i+1][j+1] = last[dp[i][j]][s[i]-'a'];}}ans[best] = '\0';printf("%s\n", ans);return 0;}