[CTSC 2012] Cheat

[题目链接]

         https://www.lydsy.com/JudgeOnline/problem.php?id=2806

[算法]

         首先建立广义后缀自动机

         注意到问题具有单调性 ， 不妨对于每组询问二分答案mid

         如何检验?

         记fi表示前i个字符最多能选几个 ， 有转移方程 ：

          fi = max{ fi - 1 , fj + i - j } (i - maxlen[i] <= j <= i - mid)

          其中maxlen[i]表示第i个字符向前最多可匹配多少个字符

          i - maxlen[i]单调递增 , i - mid同样单调递增

         单调队列优化即可

         时间复杂度 : O(NlogN)

[代码]

         

#include<bits/stdc++.h>
using namespace std;
#define N 1100010
          
int n , m , L;
int dp[N] , maxlen[N] , q[N];
char s[N];

#define rint register int

struct Suffix_Automaton
{
    int sz , last;
    int father[N] , child[N][3] , depth[N];
    Suffix_Automaton()
    {
        sz = 1;
        last = 1;
    }
    inline int new_node(int dep)
    {
        depth[++sz] = dep;
        return sz;
    }
    inline void extend(int ch)
    {
        int np = new_node(depth[last] + 1);
        int p = last;
        while (child[p][ch] == 0)
        {
            child[p][ch] = np;
            p = father[p];
        }
        if (child[p][ch] == np) father[np] = 1;
        else
        {
            int q = child[p][ch];
            if (depth[q] == depth[p] + 1) father[np] = q;
            else
            {
                int nq = new_node(depth[p] + 1);
                father[nq] = father[q];
                father[np] = father[q] = nq;
                memcpy(child[nq] , child[q] , sizeof(child[q]));
                while (child[p][ch] == q)
                {
                    child[p][ch] = nq;
                    p = father[p];
                }
            }
        }
        last = np;
      }
      inline void match()
      {
              int now = 1 , mxlen = 0;
              for (rint i = 1; i <= L; ++i)
            {
                    int nxt = s[i] - '0';
                          while (now != 1 && !child[now][nxt]) now = father[now] , mxlen = depth[now];
                if (child[now][nxt]) 
                {
                    ++mxlen;
                    now = child[now][nxt];
                } else
                {
                    mxlen = 0;
                    now = 1;
                }
                maxlen[i] = mxlen;
            }
            }
} SAM;

inline void chkmin(int &x , int y)
{
        x = min(x , y);
}
inline void chkmax(int &x , int y)
{
    x = max(x , y);
}
inline bool check(int mid)
{
    int l = 1 , r = 0;
    for (rint i = 1; i < mid; ++i) dp[i] = 0;
    for (rint i = mid; i <= L; ++i)
    {
        while (l <= r && dp[q[r]] - q[r] <= dp[i - mid] - i + mid) --r;
        q[++r] = i - mid;
        while (l <= r && q[l] < i - maxlen[i]) ++l;
        dp[i] = dp[i - 1];
        if (l <= r) chkmax(dp[i] , dp[q[l]] - q[l] + i);    
    }    
    return dp[L] * 10 >= 9 * L;
}

int main()
{
    
    scanf("%d%d" , &n , &m);
    int mxl = 0;
    for (rint i = 1; i <= m; ++i)
    {
        scanf("%s" , s + 1);
                L = strlen(s + 1);
                chkmax(mxl , L);
                for (rint j = 1; j <= L; ++j) SAM.extend(s[j] - '0');
                SAM.extend(2);
    }
    for (rint i = 1; i <= n; ++i)
    {
        scanf("%s" , s + 1);
        int l = 1 , r = strlen(s + 1) , ans = 0;
                L = r;
                chkmin(r , mxl);
        int now = 1 , mxlen = 0;
        SAM.match();
        while (l <= r)
        {
            int mid = (l + r) >> 1;
            if (check(mid))
            {
                ans = mid;
                l = mid + 1;
            } else r = mid - 1;
        }    
        printf("%d\n" , ans);
    }
    
    return 0;
}