链接:
https://vjudge.net/problem/HDU-2825
题意:
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.
思路:
先建立trie图, end记录以这个点结尾的额字符串标号, 二进制形式.
同时节点的fail指针的也要一起合并过来.
在考虑dp[i][j][k], 长度为i, 以j为结尾, 拥有k的字符串种类.
由上往下更新, trie图上下一个节点, 同时合并结尾字符串累加答案.
代码:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
//#include <memory.h>
#include <queue>
#include <set>
#include <map>
#include <algorithm>
#include <math.h>
#include <stack>
#include <string>
#include <assert.h>
#include <iomanip>
#include <iostream>
#include <sstream>
#define MINF 0x3f3f3f3f
using namespace std;
typedef long long LL;
const LL MOD = 20090717;
const LL MAXN = 110;
const int MAXASCII = 26;
struct TrieTree
{
int Next[MAXASCII];
int end;
int fail;
void Clear()
{
memset(Next, 0, sizeof(Next));
end = 0;
}
}tree[MAXN];
char s[MAXN];
LL Dp[26][110][1<<10];
int n, m, k, cnt;
void Insert(char *s, int x)
{
int len = strlen(s);
int p = 0;
for (int i = 0;i < len;i++)
{
if (tree[p].Next[s[i]-'a'] == 0)
tree[p].Next[s[i]-'a'] = ++cnt, tree[cnt].Clear();
p = tree[p].Next[s[i]-'a'];
}
tree[p].end |= 1<<x;
}
void BuildAC()
{
queue<int> que;
for (int i = 0;i < MAXASCII;i++)
{
if (tree[0].Next[i] != 0)
{
tree[tree[0].Next[i]].fail = 0;
que.push(tree[0].Next[i]);
}
}
while (!que.empty())
{
int u = que.front();
que.pop();
tree[u].end |= tree[tree[u].fail].end;
for (int i = 0;i < MAXASCII;i++)
{
if (tree[u].Next[i] != 0)
{
tree[tree[u].Next[i]].fail = tree[tree[u].fail].Next[i];
que.push(tree[u].Next[i]);
}
else
tree[u].Next[i] = tree[tree[u].fail].Next[i];
}
}
}
int GetK(int val)
{
int cn = 0;
for (int i = 0;i < m;i++)
{
if ((1<<i) & val)
cn++;
}
return cn;
}
LL Cal()
{
memset(Dp, 0, sizeof(Dp));
Dp[0][0][0] = 1;
for (int i = 0;i < n;i++)
{
for (int j = 0;j <= cnt;j++)
{
for (int w = 0; w < (1<<m); ++w)
{
if (Dp[i][j][w] == 0)
continue;
for (int t = 0;t < MAXASCII;t++)
{
Dp[i+1][tree[j].Next[t]][w|tree[tree[j].Next[t]].end] += Dp[i][j][w];
Dp[i+1][tree[j].Next[t]][w|tree[tree[j].Next[t]].end] %= MOD;
}
}
}
}
LL sum = 0;
for (int i = 0;i <= cnt;i++)
{
for (int w = 0;w < (1<<m);w++)
{
if (GetK(w) >= k)
sum = (sum+Dp[n][i][w])%MOD;
}
}
return sum;
}
int main()
{
while (~scanf("%d%d%d", &n, &m, &k))
{
if (n == 0 && m == 0 && k == 0)
break;
cnt = 0;
tree[cnt].Clear();
for (int i = 0;i < m;i++)
{
scanf("%s", s);
Insert(s, i);
}
BuildAC();
printf("%lld\n", Cal());
}
return 0;
}