題意：

給你m(m <= 10)個串，要求你構造一個長度為n 的串， 使得至少包含m 個串中的k 個。求方案數

思路：

很像那種 構造一個串  不包含m 個串的方案數那種 ， 但這種是包含。 

這個題串很小， 最多10個， 因此可以考慮狀壓dp。

那麼很好想到了。

令dp[i][j][k] 表示目前構造串的第i 位， 在自動機 j 結點，  m 個串中選擇的狀態k 的方案數。

很好轉移， 大家自己想想把。

不過寫完之後TLE了。= =

然後就加了一堆小小優化。

比如說 取模改成加減法啊， 比如說 當前一個狀態的dp 為0  就continue啊等等。（然後就608ms 過了。。 當然也寫錯了幾個小地方，不知道是不是TLE 的錯誤 。。）

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;

const int mod = 20090717;
void add(int& ans, int x){
    ans += x;
    if (ans >= mod) ans -= mod;
}

const int maxn = 100 + 1;

int bitcount(int x){
    int ans = 0;
    while(x){
        if (x & 1) ++ans;
        x>>=1;
    }
    return ans;
}

struct Trie{
    int L, root;
    int next[maxn][26];
    int fail[maxn];
    int flag[maxn];
    int sum[maxn];

    int dp[26][maxn][1024];


    void init(){
        L = 0;
        root = newnode();
    }

    int newnode(){
        for (int i = 0; i < 26; ++i){
            next[L][i] = -1;
        }
        flag[L] = 0;
        sum[L] = 0;
        return L++;
    }

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

        }
        while(!q.empty()){
            int u = q.front(); q.pop();
            sum[u] |= (flag[u] | sum[fail[u] ]);
            for (int i = 0; i < 26; ++i){

                if (next[u][i] == -1){
                    next[u][i] = next[fail[u] ][i];
                }
                else {
                    fail[next[u][i] ] = next[fail[u] ][i];
                    q.push(next[u][i]);
                }
            }


        }

    }
    void insert(char* s, int id){
        int len = strlen(s);
        int nod = root;
        for (int i = 0; i < len; ++i){
            int id = s[i] - 'a';
            if (next[nod][id] == -1){
                next[nod][id] = newnode();
            }
            nod = next[nod][id];
        }
        flag[nod] = 1 << id;
    }

    void solve(int n, int K,int m){
        for (int i = 0; i <= n; ++i){
            for (int j = 0; j < L; ++j){
                for (int k = 0; k < (1<<m); ++k){
                    dp[i][j][k] = 0;
                }
            }
        }

        dp[0][0][0] = 1;

        for (int i = 1; i <= n; ++i){
            for (int k = 0; k < (1<<m); ++k){
                for (int j = 0; j < L; ++j){
                    if (!dp[i-1][j][k]) continue;
                    for (int l = 0; l < 26; ++l){
                        int nx = next[j][l];
                        add(dp[i][nx][k | sum[nx] ], dp[i-1][j][k]);
                    }
                }
            }
        }

        int ans = 0;
        for (int j = 0; j < L; ++j){
            for (int i = 0; i < (1<<m); ++i){
                if (bitcount(i) >= K){
                    add(ans, dp[n][j][i]);
                }
            }
        }
        printf("%d\n", ans);
    }


};

int n, m, k;

char s[100];
Trie ac;
int main(){
    while(~scanf("%d %d %d",&n, &m, &k) && (n || m || k)){
        ac.init();
        for (int i = 0; i < m; ++i){
            scanf("%s", s);
            ac.insert(s, i);
        }
        ac.bfs();
        ac.solve(n, k, m);
    }
    return 0;
}

 

Wireless Password

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6832    Accepted Submission(s): 2270


Problem Description Liyuan lives in a old apartment. One day, he suddenly found that there was a wireless network in the building. Liyuan did not know the password of the network, but he got some important information from his neighbor. He knew the password consists only of lowercase letters 'a'-'z', and he knew the length of the password. Furthermore, he got a magic word set, and his neighbor told him that the password included at least k words of the magic word set (the k words in the password possibly overlapping).

For instance, say that you know that the password is 3 characters long, and the magic word set includes 'she' and 'he'. Then the possible password is only 'she'.

Liyuan wants to know whether the information is enough to reduce the number of possible passwords. To answer this, please help him write a program that determines the number of possible passwords.

Input There will be several data sets. Each data set will begin with a line with three integers n m k. n is the length of the password (1<=n<=25), m is the number of the words in the magic word set(0<=m<=10), and the number k denotes that the password included at least k words of the magic set. This is followed by m lines, each containing a word of the magic set, each word consists of between 1 and 10 lowercase letters 'a'-'z'. End of input will be marked by a line with n=0 m=0 k=0, which should not be processed.
Output For each test case, please output the number of possible passwords MOD 20090717.
Sample Input 10 2 2 hello world 4 1 1 icpc 10 0 0 0 0 0
Sample Output 2 1 14195065
Source
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