POJ 3974 Palindrome (算竞进阶习题)

hash + 二分答案

数据范围肯定不能暴力，所以考虑哈希。
把前缀和后缀都哈希过之后，扫描一边字符串，对每个字符串二分枚举回文串长度，注意要分奇数和偶数

#include <iostream>
#include <cstdio>
#define INF 0x3f3f3f3f
#define full(a, b) memset(a, b, sizeof a)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
inline int lowbit(int x){ return x & (-x); }
inline int read(){
    int X = 0, w = 0; char ch = 0;
    while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }
    while(isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
    return w ? -X : X;
}
inline int gcd(int a, int b){ return a % b ? gcd(b, a % b) : b; }
inline int lcm(int a, int b){ return a / gcd(a, b) * b; }
template<typename T>
inline T max(T x, T y, T z){ return max(max(x, y), z); }
template<typename T>
inline T min(T x, T y, T z){ return min(min(x, y), z); }
template<typename A, typename B, typename C>
inline A fpow(A x, B p, C lyd){
    A ans = 1;
    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;
    return ans;
}

const int N = 1000005;
const int P = 131;
ull f[N], p[N], rf[N];
char s[N];
int ans, n, cnt;

int solve1(int i){
    int l = 0, r = min(i - 1, n - i);
    while(l < r){
        int mid = (l + r + 1) >> 1;
        if(f[i] - f[i - mid - 1] * p[mid + 1] == rf[i] - rf[i + mid + 1] * p[mid + 1])
            l = mid;
        else r = mid - 1;
    }
    return l;
}

int solve2(int i){
    int l = 0, r = min(i - 1, n - i + 1);
    while(l < r){
        int mid = (l + r + 1) >> 1;
        if(f[i - 1] - f[i - mid - 1] * p[mid] == rf[i] - rf[i + mid] * p[mid])
            l = mid;
        else r = mid - 1;
    }
    return l;
}

int main(){

    while(1){
        scanf("%s", s + 1);
        n = strlen(s + 1);
        if(n == 3 && s[1] == 'E' && s[2] == 'N' && s[3] == 'D') break;
        full(p, 0), full(f, 0), full(rf, 0);
        ans = 0;
        p[0] = 1;
        for(int i = 1; i <= n; i ++){
            f[i] = f[i - 1] * P + (s[i] - 'a' + 1);
            p[i] = p[i - 1] * P;
        }
        for(int i = n; i >= 1; i --){
            rf[i] = rf[i + 1] * P + (s[i] - 'a' + 1);
        }
        for(int i = 1; i <= n; i ++){
            ans = max(ans, 2 * solve1(i) + 1);
            ans = max(ans, 2 * solve2(i));
        }
        printf("Case %d: %d
", ++cnt, ans);
    }
    return 0;
}