HDU 3341 (AC自动机 DP)

题目链接:点击这里

题意:给出n个DNA模式串, 和一个文本串, 调整文本串字符的顺序使得所有模式串出现的次数和最多.

因为四种字符出现的次数都有限, 可以用类似于进制的做法记录字符出现的状态, 如果四个字符的总数有a, b, c, d, 当前状态出现的次数是x, y, z, k, 那么状态就记录为x×(b+1)×(c+1)×(d+1)+y×(c+1)×(d+1)+z×(d+1)+k×1

然后建立AC自动机在上面转移状态就好了.

#include <bits/stdc++.h>
using namespace std;
#define maxn 505
#define INF 1111111

int n;
char str[55][maxn];

void change (char *s) {
int len = strlen (s);
for (int i = 0; i < len; i++) {
if (s[i] == 'A') s[i] = 'a';
else if (s[i] == 'C') s[i] = 'b';
else if (s[i] == 'G') s[i] = 'c';
else s[i] = 'd';
}
}

struct trie {
int next[maxn][4], fail[maxn], end[maxn];
int root, cnt;
int new_node () {
memset (next[cnt], -1, sizeof next[cnt]);
end[cnt++] = 0;
return cnt-1;
}
void init () {
cnt = 0;
root = new_node ();
}
void insert (char *buf) {//字典树插入一个单词
int len = strlen (buf);
int now = root;
for (int i = 0; i < len; i++) {
int id = buf[i]-'a';
if (next[now][id] == -1) {
next[now][id] = new_node ();
}
now = next[now][id];
}
end[now]++;
}
void build () {//构建fail指针
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]);
}
}
}
}
int dp[maxn][11*11*11*11];//在字典树i节点 四种字母出现状态
int bit[5];
int query (int a, int b, int c, int d, char *buf) {
memset (dp, -1, sizeof dp);
dp[0][0] = 0;
bit[0] = (b+1)*(c+1)*(d+1), bit[1] = (c+1)*(d+1), bit[2] = (d+1), bit[3] = 1;
for (int x = 0; x <= a; x++) {
for (int y = 0; y <= b; y++) {
for (int z = 0; z <= c; z++) {
for (int k = 0; k <= d; k++) {
int p = x*bit[0]+y*bit[1]+z*bit[2]+k*bit[3];
for (int i = 0; i < cnt; i++) {
if (dp[i][p] == -1)
continue;
for (int id = 0; id < 4; id++) {
if (id == 0 && x == a) continue;
if (id == 1 && y == b) continue;
if (id == 2 && z == c) continue;
if (id == 3 && k == d) continue;
int nexti = next[i][id];
int nextp = p + bit[id];
dp[nexti][nextp] = max (dp[nexti][nextp], dp[i][p] + end[nexti]);
}
}
}
}
}
}
int state = a*bit[0] + b*bit[1] + c*bit[2] + d*bit[3];
int ans = 0;
for (int i = 0; i < cnt; i++) ans = max (ans, dp[i][state]);
return ans;
}
}ac;
char buf[maxn];

int main () {
int kase = 0;
while (scanf ("%d", &n) == 1 && n) {
ac.init ();
for (int i = 1; i <= n; i++) {
scanf ("%s", str[i]);
change (str[i]);
ac.insert (str[i]);
}
ac.build ();
scanf ("%s", buf);
int len = strlen (buf);
change (buf);
int a = 0, b = 0, c = 0, d = 0;
for (int i = 0; i < len; i++) {
if (buf[i] == 'a') a++;
else if (buf[i] == 'b') b++;
else if (buf[i] == 'c') c++;
else d++;
}
printf ("Case %d: %d\n", ++kase, ac.query (a, b, c, d, buf));
}
return 0;
}