[POJ 3974] Palindrome （字符串哈希+二分）

链接

POJ 3974

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题意

给出字符串，求其最长回文子串的长度。

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思路

该题目可用manacher算法在O(n)时间内解决。
最长回文子串，也是子串+长度类型的问题，此处使用字符串哈希+二分解决。
对字符串进行正序倒序两次哈希，之后二分长度，对每个长度len枚举子串的哈希值，判断相同位置的子串正逆序哈希值是否相同，相同则代表为回文串。
复杂度O(n*log(n))。

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代码

#include <cstdio>#include <iostream>#include <cstring>#include <vector>using namespace std;typedef unsigned long long ulint;const ulint seed = 30007uLL;#define maxn 1010000ulint H[maxn], P[maxn], xp[maxn];char s[maxn];int N;void initHash(){    H[0] = s[0];    for(int i = 1; i < N; i++)        H[i] = H[i - 1]*seed + s[i];    P[N-1] = s[N-1];    for(int i = N-2; i >= 0; i--)        P[i] = P[i + 1]*seed + s[i];}ulint askHash(int l, int r){    if(l == 0) return H[r];    return H[r] - H[l - 1]*xp[r - l + 1];}ulint askP(int l, int r){    if(r == N-1) return P[l];    return P[l] - P[r + 1]*xp[r - l + 1];}bool check(int len){    //cout << "len = " << len << endl;    ulint ht, pt;    for(int i = 0, l, r; i + len - 1 < N; i++)    {        l = i, r = i + len - 1;        ht = askHash(l, r);        pt = askP(l, r);        //cout << ht << " " << pt << endl;        if(ht == pt) return true;    }    return false;}int main(){    xp[0] = 1;    for(int i = 1; i < maxn; i++)    {        xp[i] = xp[i-1] * seed;    }    int kase = 0;    while(~scanf("%s", &s))    {        if(string(s) == string("END"))            break;        N = strlen(s);        for(int i = 0; i < N; i++)        {            s[i] = s[i] - 'a' + 1;        }        initHash();/*        for(int i = 0; i < N; i++)        {            cout << askHash(i, i) << " " << askP(i, i) << endl;        }*/        vector<int> dexod, dexev;        for(int i = 1; i <= N; i++)        {            if(i % 2) dexod.push_back(i);            else dexev.push_back(i);        }        int ans = 0, l = 0, r = dexod.size() - 1, m;        while(l <= r)        {            m = (l + r) >> 1;            if(check(dexod[m]))            {                l = m + 1;                ans = dexod[m];            }            else r = m - 1;        }        l = 0, r = dexev.size() - 1;        while(l <= r)        {            m = (l + r) >> 1;            if(check(dexev[m]))            {                l = m + 1;                ans = max(ans, dexev[m]);            }            else r = m - 1;        }        printf("Case %d: ", ++kase);        cout << ans << endl;    }    return 0;}