第一次做这种题目，参考了下题解，相当于把树扯直了做DP，估计这一类题都是这个套路吧。
状态方程dp[i][next] = dp[i][next]+dp[i][j]
;dp[i][j]表示长度为i的第J个结点的时候满足题意的num，next为当前j点所能走到的下一个合法的结点。
需要用高精度，看到一些规范的高精度写法，觉得不错，有空整理下来。
不知道是不是我理解错了，按理说字符串病毒长度不应超过10.。但开到55依旧RE，开550AC。。。
1 #include <iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<algorithm>
5 #include<stdlib.h>
6 #include<vector>
7 #include<cmath>
8 #include<queue>
9 #include<set>
10 using namespace std;
11 #define N 110
12 #define LL long long
13 #define INF 0xfffffff
14 const double eps = 1e-8;
15 const double pi = acos(-1.0);
16 const double inf = ~0u>>2;
17 const int child_num = 110;
18 const int BASE = 10000;
19 const int DIG = 4;
20 char s[N*100],vir[550];
21 int id[2024];
22 struct bignum
23 {
24 int a[110],len;
25 bignum()
26 {
27 memset(a,0,sizeof(a));
28 len = 1;
29 }
30 bignum(int v)
31 {
32 memset(a,0,sizeof(a));
33 len = 0;
34 do
35 {
36 a[len++] = v%BASE;
37 v/=BASE;
38 }while(v);
39 }
40 /*bignum(const char s[])
41 {
42 memset(a,0,sizeof(a));
43 int k = strlen(s);
44 len = k/DIG;
45 if(k%DIG) len++;
46 int cnt = 0;
47 for(int i = k-1; i >= 0 ; i-=DIG)
48 {
49 int t = 0;
50 int kk = i-DIG+1;
51 if(kk<0) kk =0;
52 for(int j = kk ; j <= i ; j++)
53 t = t*10+s[j]-‘0‘;
54 a[cnt++] = t;
55 }
56 }*/
57 bignum operator + (const bignum &b)const
58 {
59 bignum res;
60 res.len = max(len,b.len);
61 int i;
62 for(i = 0 ; i < res.len ;i ++)
63 res.a[i] = 0;
64 for(i = 0 ; i < res.len ; i++)
65 {
66 res.a[i] += ((i<len)?a[i]:0)+((i<b.len)?b.a[i]:0);
67 res.a[i+1] += res.a[i]/BASE;
68 res.a[i] = res.a[i]%BASE;
69 }
70 if(res.a[res.len]>0) res.len++;
71 return res;
72 }
73 void output()
74 {
75 printf("%d",a[len-1]);
76 for(int i = len-2 ; i >=0 ; i--)
77 printf("%04d",a[i]);
78 printf("\n");
79 }
80 }dp[110][110];
81 class AC
82 {
83 private:
84 int ch[N][child_num];
85 int Q[N];
86 int val[N];
87 int fail[N];
88 //int id[N];
89 int sz;
90 public :
91 void init()
92 {
93 fail[0] = 0;
94 //for(int i = 0 ;i < child_num-32 ; i++)
95 //id[i+32] = i;
96 }
97 void reset()
98 {
99 memset(val,0,sizeof(val));
100 memset(fail,0,sizeof(fail));
101 memset(ch[0],0,sizeof(ch[0]));
102 sz = 1;
103 }
104 void insert(char *a,int key)
105 {
106 int k = strlen(a),p = 0;
107 for(int i = 0 ; i < k ;i++)
108 {
109 int d = id[a[i]];
110 if(ch[p][d]==0)
111 {
112 memset(ch[sz],0,sizeof(ch[sz]));
113 ch[p][d] = sz++;
114 }
115 p = ch[p][d];
116 }
117 val[p] = key;
118 }
119 void construct(int n)
120 {
121 int i,head=0,tail = 0;
122 for(i = 0; i < n ; i++)
123 {
124 if(ch[0][i])
125 {
126 Q[tail++] = ch[0][i];
127 fail[ch[0][i]] = 0;
128 }
129 }
130 while(head!=tail)
131 {
132 int u = Q[head++];
133 val[u]|=val[fail[u]];
134 for(i = 0 ; i < n ; i++)
135 {
136 if(ch[u][i])
137 {
138 Q[tail++] = ch[u][i];
139 fail[ch[u][i]] = ch[fail[u]][i];
140 }
141 else ch[u][i] = ch[fail[u]][i];
142 }
143 }
144 }
145 void work(int m,int n)
146 {
147 int i,j,g;
148 for(i = 1; i <= m ;i++)
149 for(j = 0 ;j <= sz; j++)
150 dp[i][j] = bignum(0);
151 dp[0][0] = bignum(1);
152 for(i = 0 ; i < m ;i++)
153 {
154 for(j = 0 ; j < sz ;j++)
155 for(g = 0 ; g < n ; g++)
156 if(!val[ch[j][g]])
157 {
158 dp[i+1][ch[j][g]]=dp[i+1][ch[j][g]]+dp[i][j];
159 }
160 }
161 bignum ans = bignum(0);
162 for(j = 0 ;j < sz ; j++)
163 ans=ans+dp[m][j];
164 ans.output();
165 }
166 }ac;
167 int main()
168 {
169 int n,m,i,p;
170 ac.init();
171 while(cin>>n>>m>>p)
172 {
173 cin>>s;
174 for(i = 0 ; i < n; i++)
175 id[s[i]] = i;
176 ac.reset();
177 for(i = 1;i <= p; i++)
178 {
179 scanf("%s",vir);
180 ac.insert(vir,1);
181 }
182 ac.construct(n);
183 ac.work(m,n);
184 }
185 return 0;
186 }
时间： 2024-10-12 00:39:40