【AC自动机】 HDOJ 2825 Wireless Password

AC自动机+状态压缩DP。用dp[i][j][k]表示走i步，到达AC自动机中的j节点，含有的字符串k种类（二进制状态压缩），所有的方案数。

#include <iostream>
#include <sstream>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <climits>
#define maxn 105
#define eps 1e-6
#define mod 20090717
#define INF 99999999
#define lowbit(x) (x&(-x))
//#define lson o<<1, L, mid
//#define rson o<<1 | 1, mid+1, R
typedef long long LL;
typedef unsigned long long ULL;
using namespace std;

struct node
{
    int next[maxn][26];
    int fail[maxn];
    int end[maxn];
    char s[maxn];
    queue<int> q;
    int top, now, root;
    int newnode(void)
    {
        end[top] = 0;
        fail[top] = -1;
        for(int i = 0; i < 26; i++)
            next[top][i] = -1;
        return top++;
    }
    void init(void)
    {
        top = 0;
        root = newnode();
    }
    void insert(int x)
    {
        int i, len = strlen(s), k;
        now = root;
        for(i = 0; i < len; i++) {
            k = s[i] - 'a';
            if(next[now][k] == -1)
                next[now][k] = newnode();
            now = next[now][k];
        }
        end[now] = x;
    }
    void build(void)
    {
        int i;
        fail[root] = root;
        now = root;
        for(i = 0; i < 26; i++)
            if(next[now][i] == -1)
                next[now][i] = root;
            else {
                fail[next[now][i]] = root;
                q.push(next[now][i]);
            }
        while(!q.empty()) {
            now = q.front();
            q.pop();
			end[now] |= end[fail[now]];
            for(i = 0; i < 26; i++)
                if(next[now][i] == -1)
                    next[now][i] = next[fail[now]][i];
                else {
                    fail[next[now][i]] = next[fail[now]][i];
                    q.push(next[now][i]);
                }
        }
    }
}trie;
int dp[30][105][1100];
int is[1100];
int n, m, kk;

void read(void)
{
    int i, tmp = 1;
    trie.init();
    for(i = 1; i <= m; i++) {
        scanf("%s", trie.s);
        trie.insert(tmp);
        tmp<<=1;
    }
    trie.build();
}
void init(void)
{
    memset(dp, 0, sizeof dp);
}
void work(void)
{
    int o = 0, i, j, k, ans, tmp, res, cnt, p;
    for(i = 0; i < m; i++) o<<=1, o+=1;
    dp[0][0][0] = 1;
    for(i = 0; i <= n; i++)
        for(j = 0; j < trie.top; j++)
            for(k = 0; k <= o; k++) {
                if(dp[i][j][k] == 0) continue;
                for(p = 0; p < 26; p++)
                    dp[i+1][trie.next[j][p]][k|trie.end[trie.next[j][p]]] = (dp[i+1][trie.next[j][p]][k|trie.end[trie.next[j][p]]] + dp[i][j][k])%mod;
            }
    ans = 0;
    cnt = 0;
    for(i = 0; i <= o; i++) {
        tmp = i;
        res = 0;
        while(tmp) res+=tmp&1, tmp>>=1;
        if(res >= kk) is[cnt++] = i;
    }
    for(i = 0; i < trie.top; i++)
        for(j = 0; j < cnt; j++)
            ans = (ans + dp[n][i][is[j]])%mod;
    printf("%d\n", ans);
}
int main(void)
{
    while(scanf("%d%d%d", &n, &m, &kk), n!=0 || m!=0 || kk!=0) {        
        init();
        read();
        work();
    }
    return 0;
}

后来对程序优化了一下，跑的更快了。

#include <iostream>
#include <sstream>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <climits>
#define maxn 105
#define eps 1e-6
#define mod 20090717
#define INF 99999999
#define lowbit(x) (x&(-x))
//#define lson o<<1, L, mid
//#define rson o<<1 | 1, mid+1, R
typedef long long LL;
typedef unsigned long long ULL;
using namespace std;

int next[maxn][26];
int fail[maxn];
int end[maxn];
char s[maxn];
int dp[30][105][1100];
int hash[1100];
int n, m, kk;
queue<int> q;
int top, now, root;

int newnode(void)
{
    end[top] = 0;
    fail[top] = -1;
    for(int i = 0; i < 26; i++)
        next[top][i] = -1;
    return top++;
}
void init(void)
{
    top = 0;
    root = newnode();
}
void insert(int x)
{
    int i, len = strlen(s), k;
    now = root;
    for(i = 0; i < len; i++) {
        k = s[i] - 'a';
        if(next[now][k] == -1)
            next[now][k] = newnode();
        now = next[now][k];
    }
    end[now] = x;
}
void build(void)
{
    int i;
    fail[root] = root;
    now = root;
    for(i = 0; i < 26; i++)
        if(next[now][i] == -1)
            next[now][i] = root;
        else {
            fail[next[now][i]] = root;
            q.push(next[now][i]);
        }
    while(!q.empty()) {
        now = q.front();
        q.pop();
        end[now] |= end[fail[now]];
        for(i = 0; i < 26; i++)
            if(next[now][i] == -1)
                next[now][i] = next[fail[now]][i];
            else {
                fail[next[now][i]] = next[fail[now]][i];
                q.push(next[now][i]);
            }
    }
}
void read(void)
{
    int i, tmp = 1;
    for(i = 1; i <= m; i++) {
        scanf("%s", s);
        insert(tmp);
        tmp<<=1;
    }
    build();
}
void work(void)
{
    int o = 0, i, j, k, ans, p;
    for(i = 0; i < m; i++) o<<=1, o+=1;
    for(i = 0; i <= n; i++)
        for(j = 0; j < top; j++)
            for(k = 0; k <= o; k++)
                dp[i][j][k] =0;
    dp[0][0][0] = 1;
    for(i = 0; i < n; i++)
        for(j = 0; j < top; j++)
            for(k = 0; k <= o; k++) {
                if(dp[i][j][k] == 0) continue;
                for(p = 0; p < 26; p++)
                    dp[i+1][next[j][p]][k|end[next[j][p]]] = (dp[i+1][next[j][p]][k|end[next[j][p]]] + dp[i][j][k])%mod;
            }
    ans = 0;
    for(i = 0; i <= o; i++)
        if(hash[i] >= kk)
            for(j = 0; j < top; j++)
                ans = (ans + dp[n][j][i])%mod;
    printf("%d\n", ans);
}
int main(void)
{
    int i, j;
    for(i = 0; i < (1<<10); i++) {
        hash[i] = 0;
        for(j = 0; j < 10; j++)
            if(i & (1<<j))
                hash[i]++;
    }
    while(scanf("%d%d%d", &n, &m, &kk), n!=0 || m!=0 || kk!=0) {        
        init();
        read();
        work();
    }
    return 0;
}