UVa 12206 (字符串哈希) Stammering Aliens

体验了一把字符串Hash的做法，感觉Hash这种人品算法好神奇。

也许这道题的正解是后缀数组，但Hash做法的优势就是编码复杂度大大降低。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int maxn = 40000 + 10;
const int x = 123;
int n, m, pos;
unsigned long long H[maxn], xp[maxn];
unsigned long long hash[maxn];
int rank[maxn];
char s[maxn];

bool cmp(const int& a, const int& b)
{ return hash[a] < hash[b] || (hash[a] == hash[b] && a < b); }

bool possible(int L)
{
    int cnt = 0;
    pos = -1;
    for(int i = 0; i + L - 1 < n; i++)
    {
        rank[i] = i;
        hash[i] = H[i] - H[i + L] * xp[L];
    }
    sort(rank, rank + n - L + 1, cmp);
    for(int i = 0; i + L - 1 < n; i++)
    {
        if(i == 0 || hash[rank[i]] != hash[rank[i - 1]]) cnt = 0;
        if(++cnt >= m) pos = max(pos, rank[i]);
    }
    return pos >= 0;
}

int main()
{
    //freopen("in.txt", "r", stdin);

    xp[0] = 1;
    for(int i = 1; i < maxn; i++) xp[i] = xp[i - 1] * x;

    while(scanf("%d", &m) == 1 && m)
    {
        scanf("%s", s);
        n = strlen(s);
        H[n] = 0;
        for(int i = n - 1; i >= 0; i--) H[i] = H[i + 1] * x + s[i] - 'a';

        if(!possible(1)) puts("none");
        else
        {
            int L = 1, R = n;
            while(L < R)
            {
                int M = (L + R + 1) / 2;
                if(possible(M)) L = M;
                else R = M - 1;
            }
            possible(L);
            printf("%d %d\n", L, pos);
        }
    }

    return 0;
}