HDU - 2825 Wireless Password(AC自动机+DP)

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 10 2 2
hello 
world 
4 1 1
icpc 
10 0 0
0 0 0 

 2
1
14195065 

题意：让你构造一个长度为n的字符串，使得至少包含共m个的集合里的k个，求个数。
思路：AC自动机构造，需要用到DP的思想，设dp[i][j][k]表示长度为i到达自动机状态为j的时候且状态k时的个数。我们需要用一个来表示状态j包含的字符串的个数，也就是走到字符串末尾的个数，二进制表示。然后还要注意一点的是：我们在构造自动机的时候有fail指针的转移，所以状态j也需要结合它失配时候的状态。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
const int mod = 20090717;

int n, m, k;
int dp[30][250][1<<10];
int num[5000];

struct Trie {
    int nxt[110][26], fail[110], end[110];
    int root, sz;

    int newNode() {
        for (int i = 0; i < 26; i++)
            nxt[sz][i] = -1;
        end[sz++] = 0;
        return sz - 1;
    }

    void init() {
        sz = 0;
        root = newNode();
    }

    void insert(char buf[], int id) {
        int now = root;
        for (int i = 0; buf[i]; i++) {
            if (nxt[now][buf[i]-‘a‘] == -1)
                nxt[now][buf[i]-‘a‘] = newNode();
            now = nxt[now][buf[i]-‘a‘];
        } 
        end[now] |= (1<<id);
    }

    void build() {
        queue<int> q;
        fail[root] = root;
        for (int i = 0; i < 26; i++) {
            if (nxt[root][i] == -1) 
                nxt[root][i] = root;
            else {
                fail[nxt[root][i]] = root;
                q.push(nxt[root][i]);
            }
        }

        while (!q.empty()) {
            int now = q.front();
            q.pop();
            end[now] |= end[fail[now]];
            for (int i = 0; i < 26; i++) {
                if (nxt[now][i] == -1)
                    nxt[now][i] = nxt[fail[now]][i];
                else {
                    fail[nxt[now][i]] = nxt[fail[now]][i];
                    q.push(nxt[now][i]);
                }
            }
        }
    }

    int solve() {
        for (int i = 0; i <= n; i++)
            for (int j = 0; j < sz; j++)
                for (int k = 0; k < (1<<m); k++)
                    dp[i][j][k] = 0;
        dp[0][0][0] = 1;
        for (int i = 0; i < n; i++)
            for (int j = 0; j < sz; j++)
                for (int k = 0; k < (1<<m); k++) 
                    if (dp[i][j][k] > 0) {
                        for (int x = 0; x < 26; x++) {
                            int ni = i+1;
                            int nj = nxt[j][x];
                            int nk = k | end[nj];
                            dp[ni][nj][nk] += dp[i][j][k];
                            dp[ni][nj][nk] %= mod;
                        }
                    }
        int ans = 0;
        for (int p = 0; p < (1<<m); p++) {
            if (num[p] < k) continue;
            for (int i = 0; i < sz; i++)
                ans = (ans + dp[n][i][p]) % mod;
        }
        return ans;
    }
} ac;
char buf[20];

int main() {
    for (int i = 0; i < (1<<10); i++) {
        num[i] = 0;
        for (int j = 0; j < 10; j++)
            if (i & (1<<j))
                num[i]++;
    }

    while (scanf("%d%d%d", &n, &m, &k) != EOF && n+m+k) {
        ac.init();
        for (int i = 0; i < m; i++) {
            scanf("%s", buf);
            ac.insert(buf, i);
        }

        ac.build();
        printf("%d\n", ac.solve());
    }
    return 0;
}

HDU - 2825 Wireless Password(AC自动机+DP)

标签：des   style   color   io   os   ar   for   sp   div   

原文地址：http://blog.csdn.net/u011345136/article/details/39810343