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。

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。

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;
}