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

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]
;
string ans[M]
;

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;
}