uvalive 4513 Stammering Aliens

题意：给你一个串，问期中至少出现m次的最长子串及其起始位置的坐标。

思路：hash+LCP+二分答案

#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];

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

int possible(int L)
{
    int c = 0;
    pos = -1;
    for(int i = 0; i < n-L+1; 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 < n-L+1; i++)
    {
        if(i == 0 || hash[rank[i]] != hash[rank[i-1]])
            c = 0;
        if(++c >= m)
            pos = max(pos, rank[i]);
    }
    return pos >= 0;
}

int main()
{
    char s[maxn];
    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');
        xp[0] = 1;
        for(int i = 1; i <= n; i++)
            xp[i] = xp[i-1]*x;
        if(!possible(1))
            printf("none
");
        else
        {
            int L = 1, R = n+1;
            while(R - L > 1)
            {
                int M = L + (R-L)/2;
                if(possible(M))
                    L = M;
                else R = M;
            }
            possible(L);
            printf("%d %d
", L, pos);
        }
    }
    return 0;
}