题目链接：hdu 2296 Ring

题目大意：给定N和M，表示要求一个长度不大于N的字符串，现在有M个得分串，给定M个得分串，以及每个串的得分

值。要求分值尽量大，字符串尽量小，字典序尽量小。

解题思路：AC自动机+DP，对得分串建立AC自动机，同样在自动机上DP，dp[i][j]表示长度i匹配到j节点的权值最大值，

rec[i][j]则表示字典序最小的字符串，过程中既要维护ans。一定要存储具体的字符串进行比较字典序，否则会WA。

#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
#include <iostream>
#include <algorithm>

using namespace std;

const int maxn = 2005;
const int sigma_size = 26;
const int inf = 0x3f3f3f3f;

struct Aho_Corasick {
    int sz, g[maxn][sigma_size];
    int tag[maxn], fail[maxn], last[maxn];

    int dp[55][maxn];
    char rec[55][maxn][55];

    void init();
    int idx(char ch);
    void insert(char* str, int k);
    void getFail();
    void match(char* str);
    void put(int x, int y);
    void solve();
}A;

int N, M;
char w[105][15];

void init () {
    A.init();

    scanf("%d%d", &N, &M);
    int x;
    for (int i = 0; i < M; i++)
        scanf("%s", w[i]);
    for (int i = 0; i < M; i++) {
        scanf("%d", &x);
        A.insert(w[i], x);
    }
}

int main () {
    int cas;
    scanf("%d", &cas);
    while (cas--) {
        init();
        A.solve();
    }
    return 0;
}

bool cmp(char* a, char* b) {
    int la = strlen(a);
    int lb = strlen(b);

    if (la != lb)
        return la > lb;
    return strcmp(a, b) > 0;
}

void Aho_Corasick::solve() {
    getFail();
    int val = 0;
    char ans[55];
    memset(ans, 0, sizeof(ans));
    memset(rec, 0, sizeof(rec));

    memset(dp, -inf, sizeof(dp));
    dp[0][0] = 0;

    for (int k = 0; k < N; k++) {

        for (int i = 0; i < sz; i++) {
            if (dp[k][i] < 0)
                continue;

            for (int j = 0; j < sigma_size; j++) {
                int u = i;
                while (u && g[u][j] == 0)
                    u = fail[u];
                u = g[u][j];

                if (dp[k+1][u] < dp[k][i] + tag[u] || (dp[k+1][u] == dp[k][i] + tag[u] && strcmp(rec[k+1][u], rec[k][i]) > 0)) {
                    dp[k+1][u] = dp[k][i] + tag[u];
                    strcpy(rec[k+1][u], rec[k][i]);
                    rec[k+1][u][k] = j + ‘a‘;

                    if (val < dp[k+1][u] || (val == dp[k+1][u] && cmp(ans, rec[k+1][u])) ) {
                        val = dp[k+1][u];
                        strcpy(ans, rec[k+1][u]);
                    }
                }
            }
        }
    }
    printf("%s\n", ans);
}

void Aho_Corasick::init() {
    sz = 1;
    tag[0] = 0;
    memset(g[0], 0, sizeof(g[0]));
}

int Aho_Corasick::idx(char ch) {
    return ch - ‘a‘;
}

void Aho_Corasick::put(int x, int y) {
}

void Aho_Corasick::insert(char* str, int k) {
    int u = 0, n = strlen(str);

    for (int i = 0; i < n; i++) {
        int v = idx(str[i]);
        if (g[u][v] == 0) {
            tag[sz] = 0;
            memset(g[sz], 0, sizeof(g[sz]));
            g[u][v] = sz++;
        }
        u = g[u][v];
    }
    //tag[u] = max(tag[u], k);
    tag[u] += k;
}

void Aho_Corasick::match(char* str) {
    int n = strlen(str), u = 0;
    for (int i = 0; i < n; i++) {
        int v = idx(str[i]);
        while (u && g[u][v] == 0)
            u = fail[u];

        u = g[u][v];

        if (tag[u])
            put(i, u);
        else if (last[u])
            put(i, last[u]);
    }
}

void Aho_Corasick::getFail() {
    queue<int> que;

    for (int i  = 0; i < sigma_size; i++) {
        int u = g[0][i];
        if (u) {
            fail[u] = last[u] = 0;
            que.push(u);
        }
    }

    while (!que.empty()) {
        int r = que.front();
        que.pop();

        for (int i = 0; i < sigma_size; i++) {
            int u = g[r][i];

            if (u == 0) {
                g[r][i] = g[fail[r]][i];
                continue;
            }

            que.push(u);
            int v = fail[r];
            while (v && g[v][i] == 0)
                v = fail[v];

            fail[u] = g[v][i];
            tag[u] += tag[fail[u]];
            //        last[u] = tag[fail[u]] ? fail[u] : last[fail[u]];
        }
    }
}

hdu 2296 Ring(AC自动机+DP)