HDU 4057 AC自动机+状压dp

题意：

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4057
要求构造一条长度为l的字符串，构造的字符串中若含有给出的子串，就可以加上该子串的权值，但是同一子串只能算一次权值，问构造的字符串最大权值多少。

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思路：

看到n只有10，典型的AC自动机状压的思路，dp[x][y][S]保存构造到第x个字符，到达结点y，且含有子串的状态为S是否可能，直接转移最后统计各种S下的最大权值即可。

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代码：

#include <bits/stdc++.h>using namespace std;const int INF = 0x3f3f3f3f;struct ACauto {    int next[1005][4], fail[1005], end[1005];    int root, sz;    int newnode() {        for (int i = 0; i < 4; i++)            next[sz][i] = -1;        end[sz++] = 0;        return sz - 1;    }    void init() {        sz = 0;        root = newnode();    }    int idx(char c) {        if (c == 'A') return 0;        if (c == 'C') return 1;        if (c == 'G') return 2;        return 3;    }    void insert(char *buf, int id) {        int len = strlen(buf);        int now = root;        for (int i = 0; i < len; i++) {            int id = idx(buf[i]);            if (next[now][id] == -1)                next[now][id] = newnode();            now = next[now][id];        }        end[now] |= (1 << id);    }    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]);                }            }        }    }} ac;int n, len;bool dp[2][1005][1100];int val[1100], v[15];const char d[] = {'A', 'C', 'G', 'T'};void solve() {    for (int S = 0; S < (1 << n); S++) {        val[S] = 0;        for (int i = 0; i < n; i++) {            if (S & (1 << i)) val[S] += v[i];        }    }    memset(dp, false, sizeof(dp));    dp[0][0][0] = true;    for (int i = 0; i < len; i++) {        for (int j = 0; j < ac.sz; j++)            for (int S = 0; S < (1 << n); S++)                dp[(i + 1) % 2][j][S] = false;        for (int j = 0; j < ac.sz; j++) {            for (int S = 0; S < (1 << n); S++) {                if (!dp[i % 2][j][S]) continue;                for (int k = 0; k < 4; k++) {                    int ni = i + 1, nj = ac.next[j][k], nS = S | ac.end[nj];                    dp[ni % 2][nj][nS] |= dp[i % 2][j][S];                }            }        }    }    int ans = -INF;    for (int j = 0; j < ac.sz; j++) {        for (int S = 0; S < (1 << n); S++) {            if (dp[len % 2][j][S])                ans = max(ans, val[S]);        }    }    if (ans < 0) puts("No Rabbit after 2012!");    else printf("%d\n", ans);}char str[1005];int main() {    //freopen("in.txt", "r", stdin);    while (scanf("%d%d", &n, &len) == 2) {        ac.init();        for (int i = 0; i < n; i++) {            scanf("%s%d", str, &v[i]);            ac.insert(str, i);        }        ac.build();        solve();    }    return 0;}