BZOJ 2580: [Usaco2012 Jan]Video Game

2580: [Usaco2012 Jan]Video Game

Time Limit: 10 Sec  Memory Limit: 128 MB
Submit: 142  Solved: 96
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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

分析：

我们定义状态f[i][j]为第i个字符匹配到第j个节点的最多匹配个数，然后求出AC自动机转移即可...

我们建出Trie图，然后转移方程就是f[i][nxt[j][k]]=max(f[i][nxt[j][k]],f[i-1][j]+num[nxt[j][k]])...

num[i]代表匹配到i号节点可以产生的匹配串个数，包括以i为结尾的串和fail指针产生的串...

代码：

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
//by NeighThorn
using namespace std;

const int maxn=15+5,maxm=1000+5;

int n,m,tot,head=0,tail=0,q[maxm],f[maxm][maxn*maxn];

char s[maxn];

struct trie{
	int cnt,fail,nxt[3];
}tr[maxm];

inline void insert(char *s){
	int len=strlen(s),p=0;
	for(int i=0;i<len;i++){
		if(!tr[p].nxt[s[i]-'A'])
			tr[p].nxt[s[i]-'A']=++tot;
		p=tr[p].nxt[s[i]-'A'];tr[p].fail=-1;
	}
	tr[p].cnt++;
}

inline void buildACM(void){
	head=0,tail=0;q[0]=0;
	while(head<=tail){
		int id=q[head++],p=-1;
		for(int i=0;i<3;i++){
			if(tr[id].nxt[i]){
				if(id){
					p=tr[id].fail;
					while(p!=-1){
						if(tr[p].nxt[i]){
							tr[tr[id].nxt[i]].fail=tr[p].nxt[i];
							break;
						}
						p=tr[p].fail;
					}
					if(p==-1) tr[tr[id].nxt[i]].fail=0;
				}
				else
					tr[tr[id].nxt[i]].fail=0;
				tr[tr[id].nxt[i]].cnt+=tr[tr[tr[id].nxt[i]].fail].cnt;
				q[++tail]=tr[id].nxt[i];
			}
			else if(id)
				tr[id].nxt[i]=tr[tr[id].fail].nxt[i];
		}
	}
}

inline void DP(void){
	memset(f,-1,sizeof(f));
	f[0][0]=0;
	for(int i=1;i<=m;i++)
		for(int j=0;j<=tot;j++)
			if(f[i-1][j]!=-1)
				for(int k=0;k<3;k++)
					if(tr[j].nxt[k])
						f[i][tr[j].nxt[k]]=max(f[i][tr[j].nxt[k]],f[i-1][j]+tr[tr[j].nxt[k]].cnt);
}

signed main(void){
	scanf("%d%d",&n,&m);tr[0].fail=-1;
	for(int i=1;i<=n;i++)
		scanf("%s",s),insert(s);
	buildACM();DP();int ans=0;
	for(int i=0;i<=tot;i++)
		ans=max(ans,f[m][i]);
	printf("%d\n",ans);
	return 0;
}

　　

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By NeighThorn