BZOJ2806CTSC2012Cheat 广义后缀自动机+二分+Dp

Posted 2020-10-29 EZ_WYC

题目

题目在这里

思路&做法

我们先对标准作文库建广义后缀自动机。

然后对于每一篇阿米巴的作文， 我们首先把放到广义后缀自动机跑一遍， 对于每一个位置， 记录公共子串的长度\((\)即代码和下文中的\(val\)数组\()\)

接着我们二分答案， 用DP检验。
Dp方程很好想， \(d_i = max \{ d_j + i - j \ | \ i-val_i <= j <= i-lim \}\)

可以用单点队列优化。

代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>

using namespace std;

const int N = 1300010; //数组不能太大， 会T的

int n, m;

int val[N];

struct Suffix_Automaton
{   int nxt[N][2], fail[N], sz;
    int len[N];
    int root;

    Suffix_Automaton() { }

    inline int newnode(int l)
    {   memset(nxt[sz], 0, sizeof(nxt[sz]));
        fail[sz] = 0;
        len[sz] = l;
        return sz++;
    }

    void init()
    {   sz = 1;
        root = newnode(0);
    }

    inline int idx(char x) { return x - ‘0‘; }

    int add(int last, char x)
    {   int c = idx(x);
        if (nxt[last][c])
        {   int p = last, q = nxt[last][c];
            if (len[q] == len[p] + 1)
                return q;
            else
            {   int u = newnode(len[p] + 1);
                for (int i = 0; i < 2; i++) nxt[u][i] = nxt[q][i];
                fail[u] = fail[q];
                fail[q] = u;
                while (p && nxt[p][c] == q)
                {   nxt[p][c] = u;
                    p = fail[p];
                }
                return u;
            }
        }
        else
        {   int now = newnode(len[last] + 1);
            int p = last;
            while (p && !nxt[p][c])
            {   nxt[p][c] = now;
                p = fail[p];
            }
            if (!p) fail[now] = root;
            else
            {   int q = nxt[p][c];
                if (len[q] == len[p] + 1)
                    fail[now] = q;
                else
                {   int u = newnode(len[p] + 1);
                    for (int i = 0; i < 2; i++) nxt[u][i] = nxt[q][i];
                    fail[u] = fail[q];
                    fail[now] = fail[q] = u;
                    while (p && nxt[p][c] == q)
                    {   nxt[p][c] = u;
                        p = fail[p];
                    }
                }
            }
            return now;
        }
    }

    void insert(char *s)
    {   int Len = strlen(s);
        int last = root;
        for (int i = 0; i < Len; i++)
            last = add(last, s[i]);
    }

    void work(char *str)
    {   int cnt = 0;
        int now = root;
        int Len = strlen(str+1);
        for (int i = 1; i <= Len; i++)
        {   int c = idx(str[i]);
            if (nxt[now][c])
            {   cnt++;
                now = nxt[now][c];
            }
            else
            {   while (now && !nxt[now][c]) now = fail[now];
                if (!now) { now = root; cnt = 0; }
                else { cnt = len[now] + 1; now = nxt[now][c]; }
            }
            val[i] = cnt;
        }
    }
} tzw;

int d[N];
int Q[N], hd, tl;
bool check(char *s, int lim)
{   int Len = strlen(s+1);
    for (int i = 0; i <= lim; i++) d[i] = 0;
    hd = 0, tl = 1;
    for (register int i = lim; i <= Len; i++)
    {   d[i] = d[i-1];
        if (i > lim) //这里一定不能去掉， 去掉会RE
        {   while (hd < tl && Q[hd] < i-val[i]) hd++;
            while (hd < tl && d[Q[tl-1]]+i-Q[tl-1] < d[i-lim]+i-(i-lim)) tl--;
            Q[tl++] = i - lim;
        }
        if (val[i] >= lim) d[i] = max(d[i], d[Q[hd]] + i - Q[hd]);
    }
    return 10*d[Len] >= 9*Len;
}

int solve(char *s)
{   int l = 1, r = strlen(s+1);
    while (l <= r)
    {   int mid = (l+r) >> 1;
        if (check(s, mid)) l = mid + 1;
        else r = mid - 1;
    }
    return r;
}

char str[N];

int main()
{   scanf("%d %d", &n, &m);
    tzw.init();
    for (register int i = 1; i <= m; i++)
    {   scanf("%s", str);
        tzw.insert(str);
    }
    for (register int i = 1; i <= n; i++)
    {   scanf("%s", str+1);
        tzw.work(str);
        if(!check(str, 1)) puts("0");
        else printf("%d\n", solve(str));
    }
    return 0;
}

备注

注释里的坑我全踩了