hdu 3068最长回文（hash）

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题目链接：【hdu 3068】

求一个字符串（len<=110000）的最长回文串

一般解法是manacher，但是这一题用hash也是可以ac的

假设当前判断的是以i为中心偶数最长回文串，那么s[2*i+1-k……i]与s[i+1……k]的哈希值必定相同

假设当前判断的是以i为中心奇数最长回文串，那么s[2*i-k……i-1]与s[i+1……k]的哈希值必定相同

用二分求出相应的k

#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>#include <string>using namespace std;#define ull unsigned long long const int inf = 110000+10;const ull bas = 311;ull lhas[inf], rhas[inf], base[inf];char s[inf];int len, ans, even, odd;int getans(int li, int x, int y){//li是可能的最长回文串的右端点 int l = y, r = li;while(l<=r){int mid = (l+r)/2;int j = x+y-mid;int right = rhas[mid] - rhas[y-1]*base[mid-y+1];int left = lhas[j] - lhas[x+1]*base[mid-y+1];if(left ^ right) r = mid-1;else l = mid+1;}return r-y+1;}int main(){base[0] = 1;for(int i=1; i<inf; i++) base[i] = base[i-1]*bas;while(~scanf("%s", s+1)){s[0] = '$';int len = strlen(s+1);lhas[len+1] = rhas[0] = 0;for(int i=1; i<=len; i++){rhas[i] = rhas[i-1]*bas+s[i]-'a';}for(int i=len; i>=1; i--){lhas[i] = lhas[i+1]*bas+s[i]-'a';}ans = 1;for(int i=1; i<=len; i++){if(s[i]==s[i+1]){even = getans(min(2*i, len), i, i+1);ans = max(ans, 2*even);}if(s[i-1]==s[i+1]){odd = getans(min(2*i-1, len), i-1, i+1);ans = max(ans, 2*odd+1); }}printf("%d\n", ans);} return 0;}

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