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一.题目链接
POJ-3974
二.题目大意:
求 s 的最大回文子串长度.
三.分析:
此题可以用 字符串 Hash + 二分 或者是 Manacher 算法.(后者明显比前者块)
由于这是 Manacher 的模板题,所以这里只讲第一种方法.
O(N)扫描字符串 s,建立前缀与后缀 Hash 数组.
之后枚举回文串的中心位置,二分答案即可.
ps:要分别处理奇偶长度的回文串.
的算法,Manacher 的时间复杂度为 ,但用时 Manacher 完胜第一种.
四.代码实现:
字符串 Hash + 二分
#include <set>
#include <map>
#include <ctime>
#include <queue>
#include <cmath>
#include <stack>
#include <bitset>
#include <vector>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#define eps 1e-8
#define lc k * 2
#define rc k * 2 + 1
#define pi acos(-1.0)
#define ll long long
#define ull unsigned long long
using namespace std;const int M = (int)1e6;
const int mod = 99991;
const int inf = 0x3f3f3f3f;char s[M + 5];
ull P[M + 5];
ull prefix[M + 5];
ull suffix[M + 5];void get_sum(int len)
{P[0] = 1;suffix[len + 1] = 0;for(int i = 1; i <= len; ++i){prefix[i] = prefix[i - 1] * 131 + s[i] - 'a' + 1;P[i] = P[i - 1] * 131;}for(int i = len; i; --i)suffix[i] = suffix[i + 1] * 131 + s[i] - 'a' + 1;
}ull get_prefix(int l, int r)
{return prefix[r] - prefix[l - 1] * P[r - l + 1];
}ull get_suffix(int l, int r)
{return suffix[l] - suffix[r + 1] * P[r - l + 1];
}int solve()
{int len = strlen(s + 1);get_sum(len);int ans = 0, l, r, mid;for(int i = 1; i <= len; ++i){l = 0;r = min(i - 1, len - i);while(l < r){mid = (l + r + 1) >> 1;if(get_prefix(i - mid, i - 1) == get_suffix(i + 1, i + mid))l = mid;elser = mid - 1;}ans = max(ans, 2 * r + 1);l = 0;r = min(i, len - i);while(l < r){mid = (l + r + 1) >> 1;if(get_prefix(i - mid + 1, i) == get_suffix(i + 1, i + mid))l = mid;elser = mid - 1;}ans = max(ans, r * 2);}return ans;
}int main()
{int ca = 0;while(~scanf("%s", s + 1) && strcmp(s + 1, "END"))printf("Case %d: %d\n", ++ca, solve());return 0;
}
Manacher
#include <set>
#include <map>
#include <ctime>
#include <queue>
#include <cmath>
#include <stack>
#include <bitset>
#include <vector>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#define eps 1e-8
#define lc k * 2
#define rc k * 2 + 1
#define pi acos(-1.0)
#define ll long long
#define ull unsigned long long
using namespace std;const int M = (int)1e6;
const int mod = 99991;
const int inf = 0x3f3f3f3f;char s[M + 5];
int P[2 * M + 5];
char res[2 * M + 5];int Manacher()
{int len = strlen(s);res[0] = '$', res[1] = '#';for(int i = 0; i < len; ++i){res[(i + 1) * 2] = s[i];res[(i + 1) * 2 + 1] = '#';}len = 2 * len + 1;res[len + 1] = '\0';int right = 0, mid = 0;int MaxLen = 0, MaxMid = 0;for(int i = 1; i <= len; ++i){P[i] = (right > i ? min(P[2 * mid - i], right - i) : 1);while(res[i + P[i]] == res[i - P[i]])P[i]++;if(right < i + P[i]){mid = i;right = i + P[i];}if(MaxLen < P[i]){MaxMid = i;MaxLen = P[i];}}return MaxLen - 1;
}int main()
{int ca = 0;while(~scanf("%s", s) && strcmp(s, "END"))printf("Case %d: %d\n", ++ca, Manacher());return 0;
}
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