HDU2296——Ring（AC自动机+DP）

题意：输入N代表字符串长度，输入M代表喜欢的词语的个数，接下来是M个词语，然后是M个词语每个的价值。求字符串的最大价值。每个单词的价值就是单价*出现次数。单词可以重叠。如果不止一个答案，选择字典序最小的。

题解：AC自动机+dp。dp[i][j]表示在字符串长度i，在自动机的第j个状态。因为要字典序最小，所以转移时要保持字典序最小。

想了各种转移姿势 最后还是查了题解 发现可以直接记录前缀转移……

#include <bits/stdc++.h>

using namespace std;

const int N = 1500;
const int A = 26;
const int M = 105;

struct ACAutomata {

    int next[N][A], fail[N], end[N];
    int root, L;
    int alp[N];

    int idx(char ch)
    {
        return ch - ‘a‘;
    }
    int newNode()
    {
        for (int i = 0; i < A; ++i) next[L][i] = -1;
        end[L] = 0;
        return L++;
    }
    void init()
    {
        L = 0;
        root = newNode();
    }
    void insert(char buf[], int v)
    {
        int len = strlen(buf);
        int now = root;
        for (int i = 0; i < len; ++i) {
            int ch = idx(buf[i]);
            if (next[now][ch] == -1) {
                next[now][ch] = newNode();
                alp[L-1] = ch;
            }
            now = next[now][ch];
        }
        end[now] += v;
    }
    void build()
    {
        queue<int> Q;
        for (int i = 0; i < A; ++i) {
            if (next[root][i] == -1) {
                next[root][i] = root;
            } else {
                fail[ next[root][i] ] = root;
                Q.push( next[root][i] );
            }
        }
        while (Q.size()) {
            int now = Q.front();
            Q.pop();
            end[now] += end[ fail[now] ]; //注意这里！
            for (int i = 0; i < A; ++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]);
                }
            }
        }
    }

} ac;

char buf[M][20];
int v[M];

int dp[M][N];
string ans[M][N];

int main()
{
    int T;
    scanf("%d", &T);
    while (T--) {
        int n, m;
        scanf("%d%d", &n, &m);
        ac.init();
        for (int i = 0; i < m; ++i) scanf("%s", buf[i]);
        for (int i = 0; i < m; ++i) scanf("%d", &v[i]);
        for (int i = 0; i < m; ++i) ac.insert(buf[i], v[i]);
        ac.build();
        memset(dp, -1, sizeof dp);
        dp[0][0] = 0;
        ans[0][0] = "";
        for (int i = 0; i < n; ++i)
            for (int j = 0; j < ac.L; ++j)
                if (dp[i][j] >= 0)
                    for (int k = 0; k < 26; ++k) {
                        int nt = ac.next[j][k];
                        if (dp[i][j]+ac.end[nt] > dp[i+1][nt] || dp[i][j]+ac.end[nt] == dp[i+1][nt] && ans[i][j]+char(k+‘a‘) < ans[i+1][nt]) {
                            dp[i+1][nt] = dp[i][j] + ac.end[nt];
                            ans[i+1][nt] = ans[i][j] + char(k+‘a‘);
                        }
                    }

        string res;
        int maxv = 0;
        for (int i = 0; i <= n; ++i)
            for (int j = 0; j < ac.L; ++j)
                if (maxv < dp[i][j]) { maxv = dp[i][j]; res = ans[i][j]; }
        for (int i = 0; i <= n; ++i)
            for (int j = 0; j < ac.L; ++j)
                if (maxv == dp[i][j])
                    if (ans[i][j].size() < res.size() || ans[i][j].size() == res.size() && ans[i][j] < res) res = ans[i][j];

        cout << res << endl;
    }
    return 0;
}