【洛谷 P3041】 [USACO12JAN]视频游戏的连击Video Game Combos（AC自动机，dp）

题目链接

手写一下AC自动机（我可没说我之前不是手写的）

Trie上dp，每个点的贡献加上所有是他后缀的串的贡献，也就是这个点到根的fail链的和。

#include <cstdio>
#include <queue>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAXK = 1010;
const int MAXN = 1010;
struct ACA{
	int next[5], num, fail;
}AC[MAXN];
int f[MAXK][MAXN];
int n, k, cnt, ans;
char a[30];
queue <int> q;
void insert(){
	int len = strlen(a + 1), p = 0, w;
	for(int i = 1; i <= len; ++i){
	   w = a[i] - 'A';
	   if(!AC[p].next[w])
	     AC[p].next[w] = ++cnt;
	   p = AC[p].next[w];
	}
	++AC[p].num;
}
void build(){
    int now;
    for(int i = 0; i < 3; ++i)
       if(AC[0].next[i])
         q.push(AC[0].next[i]);
    while(q.size()){
        now = q.front(); q.pop();
        for(int i = 0; i < 3; ++i)
           if(AC[now].next[i]){
             AC[AC[now].next[i]].fail = AC[AC[now].fail].next[i];
             q.push(AC[now].next[i]);
           }
           else AC[now].next[i] = AC[AC[now].fail].next[i];
        AC[now].num += AC[AC[now].fail].num;
    }
}
void dp(){
    for(int i = 0; i <= k; ++i)
       for(int j = 1; j <= cnt; ++j)
          f[i][j] = -2147483647;
    for(int i = 1; i <= k; ++i)
       for(int j = 0; j <= cnt; ++j){
           for(int k = 0; k < 3; ++k)
              f[i][AC[j].next[k]] = max(f[i][AC[j].next[k]], f[i - 1][j] + AC[AC[j].next[k]].num);
       }
    for(int i = 1; i <= cnt; ++i)
       ans = max(ans, f[k][i]);
}
int main(){
	scanf("%d%d", &n, &k);
	for(int i = 1; i <= n; ++i){
		scanf("%s", a + 1);
		insert();
	}
	build();
	dp();
	printf("%d\n", ans);
	return 0;
}