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 ;
}