BZOJ 1212 L语言(DP+字典树)

求能被理解的最长前缀。

很显然的dp。令dp[i]=true,表示前缀i能理解。否则不能理解。那么dp[i+len]=dp[i]=true,当s[len]能匹配str[i,i+len].

由于模式串长度为10.且匹配过程可以用字典树加速。

所以复杂度就是O(10*m*len).

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi acos(-1.0)
# define eps 1e-9
# define MOD 1000000000
# define INF 1000000000
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<1,l,mid
# define rch p<<1|1,mid+1,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
int Scan() {
    int x=0,f=1;char ch=getchar();
    while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
    while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
    return x*f;
}
void Out(int a) {
    if(a<0) {putchar(‘-‘); a=-a;}
    if(a>=10) Out(a/10);
    putchar(a%10+‘0‘);
}
const int N=1000005;
//Code begin...

int trie[205][27], top;
char str[N], s[15];
bool vis[N];

void init(){top=1; mem(trie[0],0);}
void ins(char *s){
    int rt, nxt;
    for (rt=0; *s; rt=nxt, ++s){
        nxt=trie[rt][*s-‘a‘];
        if (!nxt) mem(trie[top],0), trie[rt][*s-‘a‘]=nxt=top++;
    }
    trie[rt][26]=1;
}
void find(int l, int r){
    int rt, nxt, i;
    for (rt=0, i=l; i<=r; rt=nxt, ++i) {
        nxt=trie[rt][str[i]-‘a‘];
        if (!nxt) return ;
        if (trie[nxt][26]) vis[i]=true;
    }
}
int main ()
{
    int n, m;
    scanf("%d%d",&n,&m); init();
    FOR(i,1,n) scanf("%s",s+1), ins(s+1);
    FOR(i,1,m) {
        scanf("%s",str+1);
        int len=strlen(str+1);
        mem(vis,0); vis[0]=true;
        int ans;
        FOR(j,0,len) if (vis[j]) find(j+1>len?len:j+1,j+10>len?len:j+10), ans=j;
        printf("%d\n",ans);
    }
    return 0;
}