Lost's revenge HDU

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("******
")
#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 = 13331;
const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
const int maxn = 1e6 + 7;
const int maxm = 8e6 + 10;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;

int n, dp[510][11 * 11 * 11 * 11 + 10], num[5], bit[5];
char buf[50];

int get_num(char ch) {
    if (ch == 'A') return 0;
    if (ch == 'T') return 1;
    if (ch == 'C') return 2;
    if (ch == 'G') return 3;
}

struct Aho_Corasick {
    int next[510][4], fail[510], End[510];
    int root, cnt;

    int newnode() {
        for (int i = 0; i < 4; i++) next[cnt][i] = -1;
        End[cnt++] = 0;
        return cnt - 1;
    }

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

    void insert(char buf[]) {
        int len = strlen(buf);
        int now = root;
        for (int i = 0; i < len; i++) {
            if (next[now][get_num(buf[i])] == -1) 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 = 0; i < 4; 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 now = Q.front();
            Q.pop();
            End[now] += End[fail[now]];
            for (int i = 0; i < 4; i++)
                if (next[now][i] == -1) 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, 0);
        for (int i = 0; i < len; ++i) num[get_num(buf[i])]++;
        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);
        mem(dp, -1);
        dp[0][0] = 0;
        for (int A = 0; A <= num[0]; ++A) {
            for (int B = 0; B <= num[1]; ++B) {
                for (int C = 0; C <= num[2]; ++C) {
                    for (int D = 0; D <= num[3]; ++D) {
                        for (int i = 0; i < cnt; ++i) {
                            int s = A * bit[0] + B * bit[1] + C * bit[2] + D * bit[3];
                            if (dp[i][s] == -1) continue;
                            for (int k = 0; k < 4; ++k) {
                                if (k == 0 && A == num[0]) continue;
                                if (k == 1 && B == num[1]) continue;
                                if (k == 2 && C == num[2]) continue;
                                if (k == 3 && D == num[3]) continue;
                                int idx = next[i][k];
                                dp[idx][s + bit[k]] = max(dp[idx][s + bit[k]], dp[i][s] + End[idx]);
                            }
                        }
                    }
                }
            }
        }
        int ans = 0, status = num[0] * bit[0] + num[1] * bit[1] + num[2] * bit[2] + num[3] * bit[3];
        for (int i = 0; i < cnt; ++i) ans = max(ans, dp[i][status]);
        return ans;
    }


    void debug() {
        for (int i = 0; i < cnt; i++) {
            printf("id = %3d,fail = %3d,end = %3d,chi = [", i, fail[i], End[i]);
            for (int j = 0; j < 26; j++) printf("%2d", next[i][j]);
            printf("]
");
        }
    }
} ac;

int main() {
    //FIN;
    int cas = 1;
    while (sfi(n) && n) {
        ac.init();
        for (int i = 0; i < n; ++i) {
            sfs(buf);
            ac.insert(buf);
        }
        ac.build();
        sfs(buf);
        printf("Case %d: %d
", cas++, ac.solve(buf));
    }
    return 0;
}