Ring HDU

题意：

就是现在给出m个串，每个串都有一个权值，现在你要找到一个长度不超过n的字符串，

其中之前的m个串每出现一次就算一次那个字符串的权值，

求能找到的最大权值的字符串，如果存在多个解，输出最短的字典序最小的串。

当最大全权值为0时输出空串。

输入最多100个子串，权值为不超过100的正整数。

每个子串长度至少为1，不超过10, n <= 50

如果不考虑方案输出，这题就变得相当简单了。

dp【i】【j】表示走到长度为 i 的时候 ，到AC自动机 j 这个节点所获得的最大权值和。

我一开始的做法是在dp的过程中获得那个最优方案。

最后死活过不去，就换了一种超级暴力的写法。

将所有情况保存下来，去一个个找字典序最小的方案。

#include <set>
#include <map>
#include <stack>
#include <queue>
#include <cmath>
#include <cstdio>
#include <string>
#include <vector>
#include <time.h>
#include <cstring>
#include <iostream>
#include <algorithm>

#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  sf(n)             scanf("%d", &n)
#define  sff(a, b)         scanf("%d %d", &a, &b)
#define  sfff(a, b, c)     scanf("%d %d %d", &a, &b, &c)
#define  sffff(a, b, c, d) scanf("%d %d %d %d", &a, &b, &c, &d)
#define  pf                printf
#define  FIN               freopen("../date.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 int maxn = 1e6 + 7;
const int maxm = 8e6 + 10;
const int INF = 0x3f3f3f3f;
const int mod = 20090717;


char str[105][12];
int n, m, cost[105];


struct Aho_Corasick {
    int next[1010][26], fail[1010], End[1010];
    int root, cnt;

    int newnode() {
        for (int i = 0; i < 26; i++) next[cnt][i] = -1;
        End[cnt++] = 0;
        return cnt - 1;
    }

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

    void insert(char buf[], int id) {
        int len = strlen(buf);
        int now = root;
        for (int i = 0; i < len; i++) {
            if (next[now][buf[i] - 'a'] == -1) next[now][buf[i] - 'a'] = newnode();
            now = next[now][buf[i] - 'a'];
        }
        End[now] += id;
    }

    void build() {
        queue<int> Q;
        fail[root] = root;
        for (int i = 0; i < 26; 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 < 26; 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]);
                }
        }
    }

    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 dp[55][1010];
string path[55][1010];

int main() {
  // FIN;
    int T;
    sf(T);
    while (T--) {
        sff(n, m);
        for (int i = 0; i < m; ++i) scanf("%s", str[i]);
        for (int i = 0; i < m; i++) sf(cost[i]);
        ac.init();
        for (int i = 0; i < m; ++i) ac.insert(str[i], cost[i]);
        ac.build();
        mem(dp, -1);
        for (int i = 0; i <= n; i++)
            for (int j = 0; j < ac.cnt; j++)
                path[i][j] = "";
        int ans = 0, num1 = 0, num2 = 0;
        dp[0][0] = 0;
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < ac.cnt; ++j) {
                if (dp[i][j] == -1) continue;
                for (int k = 0; k < 26; ++k) {
                    int idx = ac.next[j][k];
                    if (dp[i + 1][idx] < dp[i][j] + ac.End[idx]) {
                        dp[i + 1][idx] = dp[i][j] + ac.End[idx];
                        path[i + 1][idx] = path[i][j] + char(k + 'a');
                    } else if (dp[i + 1][idx] == dp[i][j] + ac.End[idx] &&
                               path[i + 1][idx] > path[i][j] + char(k + 'a')) {
                        //fuck(path[i][j]);
                        path[i + 1][idx] = path[i][j] + char(k + 'a');
                      //  fuck(path[i][j]);

                    }
                }
            }
        }
        for (int i = 1; i <= n; i++) {
            for (int j = 0; j < ac.cnt; ++j) {
                if (ans < dp[i][j]) {
                    ans = dp[i][j], num1 = i, num2 = j;
                } else if (ans == dp[i][j]&&path[num1][num2] > path[i][j] && path[num1][num2].size()>=path[i][j].size()) {
                    num1 = i, num2 = j;
                }
            }
        }
//        printf("%d
", ans);
        cout << path[num1][num2] << endl;
    }
    return 0;
}