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

4

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 = 301, B = 3;
int T, K;
char s[17];
int ch[MAXN][4], f[1001][MAXN], fail[MAXN], val[MAXN], tot = 0, root = 0;
void insert(char *s) {
    int N = strlen(s + 1);
    int now = root; 
    for(int i = 1; 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 = 0; i < B; i++) if(ch[root][i]) q.push(ch[root][i]);
    while(!q.empty()) {
        int p = q.front(); q.pop();
        for(int i = 0; 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 = 0; i <= K; i++) f[i][0] = 0;//óDD?×′ì?ê?2??é?ü′?μ?μ?￡?òò′?Dèòa?e2??üD? 
    int ans = 0;
    for(int i = 1; i <= K; i++) 
        for(int j = 0; j <= tot; j++) 
            for(int k = 0; k < B; k++) {
                int son = ch[j][k];
                if(son) {
                    f[i][son] = max(f[i][son], f[i - 1][j] + val[son]);
                    printf("%d %d %d %d %d\n", i, j, k, son, f[i][son]); 
                } 
                    
            }
    for(int i = 0; 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 = 1; i <= T; i++) 
        scanf("%s", s + 1), insert(s);
    GetFail();
    printf("%d", Dp());
    return 0;
}