Lost's revenge HDU - 3341 AC自动机+DP(需要学会如何优雅的压缩状态)

题意：

给你n个子串和一个母串，让你重排母串最多能得到多少个子串出现在重排后的母串中。

首先第一步肯定是获取母串中每个字母出现的次数，只有A T C G四种。

这个很容易想到一个dp状态dp【i】【A】【B】【C】【D】

表示在AC自动机 i 这个节点上，用了A个A，B个T，C个C，D个G。

然后我算了一下内存，根本开不下这么大的内存。

看了网上题解，然后用通过状压把,A,B,C,D压缩成一维。

这个状压就是通过进制实现需要实现唯一表示

bit[0] = 1;

bit[1] = (num[0] + 1);

bit[2] = (num[0] + 1) * (num[1] + 1);

bit[3] = (num[0] + 1) * (num[1] + 1) * (num[2] + 1);

这样就实现了A,B,C,D的唯一表示。

知道如何优化空间这题就变得非常简单了。

 #include <set>
 #include <map>
 #include <stack>
 #include <queue>
 #include <cmath>
 #include <ctime>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 #include <unordered_map>

 #define  pi    acos(-1.0)
 #define  eps   1e-9
 #define  fi    first
 #define  se    second
 #define  rtl   rt<<1
 #define  rtr   rt<<1|1
 #define  bug                printf("******\n")
 #define  mem(a, b)          memset(a,b,sizeof(a))
 #define  name2str(x)        #x
 #define  fuck(x)            cout<<#x" = "<<x<<endl
 #define  sfi(a)             scanf("%d", &a)
 #define  sffi(a, b)         scanf("%d %d", &a, &b)
 #define  sfffi(a, b, c)     scanf("%d %d %d", &a, &b, &c)
 #define  sffffi(a, b, c, d) scanf("%d %d %d %d", &a, &b, &c, &d)
 #define  sfL(a)             scanf("%lld", &a)
 #define  sffL(a, b)         scanf("%lld %lld", &a, &b)
 #define  sfffL(a, b, c)     scanf("%lld %lld %lld", &a, &b, &c)
 #define  sffffL(a, b, c, d) scanf("%lld %lld %lld %lld", &a, &b, &c, &d)
 #define  sfs(a)             scanf("%s", a)
 #define  sffs(a, b)         scanf("%s %s", a, b)
 #define  sfffs(a, b, c)     scanf("%s %s %s", a, b, c)
 #define  sffffs(a, b, c, d) scanf("%s %s %s %s", a, b,c, d)
 #define  FIN                freopen("../in.txt","r",stdin)
 #define  gcd(a, b)          __gcd(a,b)
 #define  lowbit(x)          x&-x
 #define  IO                 iOS::sync_with_stdio(false)

 using namespace std;
 typedef long long LL;
 typedef unsigned long long ULL;
 const ULL seed = ;
 const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
 const int maxn = 1e6 + ;
 const int maxm = 8e6 + ;
 const int INF = 0x3f3f3f3f;
 const int mod = 1e9 + ;

 int n, dp[][ *  *  *  + ], num[], bit[];
 char buf[];

 int get_num(char ch) {
     if (ch == 'A') return ;
     if (ch == 'T') return ;
     if (ch == 'C') return ;
     if (ch == 'G') return ;
 }

 struct Aho_Corasick {
     int next[][], fail[], End[];
     int root, cnt;

     int newnode() {
         for (int i = ; i < ; i++) next[cnt][i] = -;
         End[cnt++] = ;
         return cnt - ;
     }

     void init() {
         cnt = ;
         root = newnode();
     }

     void insert(char buf[]) {
         int len = strlen(buf);
         int now = root;
         for (int i = ; i < len; i++) {
             if (next[now][get_num(buf[i])] == -) next[now][get_num(buf[i])] = newnode();
             now = next[now][get_num(buf[i])];
         }
         End[now]++;
     }

     void build() {
         queue<int> Q;
         fail[root] = root;
         for (int i = ; i < ; i++)
             if (next[root][i] == -) next[root][i] = root;
             else {
                 fail[next[root][i]] = root;
                 Q.push(next[root][i]);
             }
         while (!Q.empty()) {
             int now = Q.front();
             Q.pop();
             End[now] += End[fail[now]];
             for (int i = ; i < ; i++)
                 if (next[now][i] == -) next[now][i] = next[fail[now]][i];
                 else {
                     fail[next[now][i]] = next[fail[now]][i];
                     Q.push(next[now][i]);
                 }
         }
     }

     int solve(char buf[]) {
         int len = strlen(buf);
         mem(num, );
         for (int i = ; i < len; ++i) num[get_num(buf[i])]++;
         bit[] = ;
         bit[] = (num[] + );
         bit[] = (num[] + ) * (num[] + );
         bit[] = (num[] + ) * (num[] + ) * (num[] + );
         mem(dp, -);
         dp[][] = ;
         for (int A = ; A <= num[]; ++A) {
             for (int B = ; B <= num[]; ++B) {
                 for (int C = ; C <= num[]; ++C) {
                     for (int D = ; D <= num[]; ++D) {
                         for (int i = ; i < cnt; ++i) {
                             int s = A * bit[] + B * bit[] + C * bit[] + D * bit[];
                             if (dp[i][s] == -) continue;
                             for (int k = ; k < ; ++k) {
                                 if (k ==  && A == num[]) continue;
                                 if (k ==  && B == num[]) continue;
                                 if (k ==  && C == num[]) continue;
                                 if (k ==  && D == num[]) continue;
                                 int idx = next[i][k];
                                 dp[idx][s + bit[k]] = max(dp[idx][s + bit[k]], dp[i][s] + End[idx]);
                             }
                         }
                     }
                 }
             }
         }
         int ans = , status = num[] * bit[] + num[] * bit[] + num[] * bit[] + num[] * bit[];
         for (int i = ; i < cnt; ++i) ans = max(ans, dp[i][status]);
         return ans;
     }

     void debug() {
         for (int i = ; i < cnt; i++) {
             printf("id = %3d,fail = %3d,end = %3d,chi = [", i, fail[i], End[i]);
             for (int j = ; j < ; j++) printf("%2d", next[i][j]);
             printf("]\n");
         }
     }
 } ac;

 int main() {
     //FIN;
     int cas = ;
     while (sfi(n) && n) {
         ac.init();
         for (int i = ; i < n; ++i) {
             sfs(buf);
             ac.insert(buf);
         }
         ac.build();
         sfs(buf);
         printf("Case %d: %d\n", cas++, ac.solve(buf));
     }
     return ;
 }