BZOJ2580: [Usaco2012 Jan]Video Game(AC自动机)

Time Limit: 10 Sec Memory Limit: 128 MB
Submit: 159 Solved: 110
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Description

Bessie is playing a video game! In the game, the three letters 'A', 'B', and 'C' are the only valid buttons. Bessie may press the buttons in any order she likes; however, there are only N distinct combos possible (1 <= N <= 20). Combo i is represented as a string S_i which has a length between 1 and 15 and contains only the letters 'A', 'B', and 'C'. Whenever Bessie presses a combination of letters that matches with a combo, she gets one point for the combo. Combos may overlap with each other or even finish at the same time! For example if N = 3 and the three possible combos are "ABA", "CB", and "ABACB", and Bessie presses "ABACB", she will end with 3 points. Bessie may score points for a single combo more than once. Bessie of course wants to earn points as quickly as possible. If she presses exactly K buttons (1 <= K <= 1,000), what is the maximum number of points she can earn?

给出n个ABC串combo[1..n]和k，现要求生成一个长k的字符串S，问S与word[1..n]的最大匹配数

Input

Line 1: Two space-separated integers: N and K. * Lines 2..N+1: Line i+1 contains only the string S_i, representing combo i.

Output

Line 1: A single integer, the maximum number of points Bessie can obtain.

Sample Input

3 7 ABA CB ABACB

Sample Output

HINT

The optimal sequence of buttons in this case is ABACBCB, which gives 4 points--1 from ABA, 1 from ABACB, and 2 from CB.

Source

AC自动机应该不难看出来

按照套路dp，设$f[i][j]$表示枚举到第$i$个位置，现在位于自动机上的第$i$位。

转移的时候枚举下一个位置就好

有两个需要注意的地方

1.Trie树在我们建fail树的时候实际被我们改造成了Trie图，因此每个节点是可能被多次枚举到的，需要对自身取$max$

2.有些深度大于当前枚举长度的点是不可能走到的，因此开始时应把每个点的权值设为$-INF$（root除外）

// luogu-judger-enable-o2

// luogu-judger-enable-o2

#include<cstdio>

#include<cstring>

#include<queue>

using namespace std;

const int MAXN = , B = ;

int T, K;

char s[];

int ch[MAXN][], f[][MAXN], fail[MAXN], val[MAXN], tot = , root = ;

void insert(char *s) {

    int N = strlen(s + );

    int now = root;

    for(int i = ; i <= N; i++) {

        int x = s[i] - 'A';

        if(!ch[now][x]) ch[now][x] = ++tot;

        now = ch[now][x];

    }

    val[now]++;

}

void GetFail() {

    queue<int> q;

    for(int i = ; i < B; i++) if(ch[root][i]) q.push(ch[root][i]);

    while(!q.empty()) {

        int p = q.front(); q.pop();

        for(int i = ; i < B; i++) {

            if(!ch[p][i]) ch[p][i] = ch[fail[p]][i];

            else fail[ch[p][i]] = ch[fail[p]][i], q.push(ch[p][i]);

        }

        val[p] += val[fail[p]];

    }

}

int Dp() {

    memset(f, -0x3f, sizeof(f));

    for(int i = ; i <= K; i++) f[i][] = ;//óDD?×′ì?ê?2??é?ü′?μ?μ?￡?òò′?Dèòa?e2??üD?

    int ans = ;

    for(int i = ; i <= K; i++)

        for(int j = ; j <= tot; j++)

            for(int k = ; k < B; k++) {

                int son = ch[j][k];

                if(son) {

                    f[i][son] = max(f[i][son], f[i - ][j] + val[son]);

                    printf("%d %d %d %d %d\n", i, j, k, son, f[i][son]);

                } 

            }

    for(int i = ; i <= tot; i++)

        ans = max(ans, f[K][i]);

    return ans;

}

int main() {

#ifdef WIN32

    freopen("a.in", "r", stdin);

#endif

    scanf("%d %d", &T, &K);

    for(int i = ; i <= T; i++)

        scanf("%s", s + ), insert(s);

    GetFail();

    printf("%d", Dp());

    return ;

}