题目大意;给定N个DNA序列,问说最少多长的字符串包含所有序列。
解题思路:AC自动机+状压DP,先对字符串构造AC自动机,然后在dp[s][i]表示匹配了s,移动到节点i时候的最短步数。
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
typedef pair<int,int> pii;
const int maxn = 205;
const int sigma_size = 4;
const int inf = 0x3f3f3f3f;
struct Aho_Corasick {
int sz, g[maxn][sigma_size];
int tag[maxn], fail[maxn], last[maxn];
int dp[maxn][1030];
void init();
int idx(char ch);
void insert(char* str, int k);
void getFail();
void match(char* str);
void put(int x, int y);
int solve();
}AC;
int N;
char w[30];
int main () {
int cas;
scanf("%d", &cas);
while (cas--) {
AC.init();
scanf("%d", &N);
for (int i = 0; i < N; i++) {
scanf("%s", w);
AC.insert(w, i);
}
printf("%d\n", AC.solve());
}
return 0;
}
int Aho_Corasick::solve() {
getFail();
memset(dp, inf, sizeof(dp));
dp[0][0] = 0;
queue<pii> que;
que.push(make_pair(0, 0));
while(!que.empty()) {
int u = que.front().first;
int s = que.front().second;
que.pop();
for (int i = 0; i < 4; i++) {
int k = u;
while (k && g[k][i] == 0)
k = fail[k];
k = g[k][i];
int ss = s | tag[k];
if (dp[k][ss] > dp[u][s] + 1) {
dp[k][ss] = dp[u][s] + 1;
que.push(make_pair(k, ss));
if (ss == (1<<N)-1)
return dp[k][ss];
}
}
}
return 0;
}
void Aho_Corasick::init() {
sz = 1;
tag[0] = 0;
memset(g[0], 0, sizeof(g[0]));
}
int Aho_Corasick::idx(char ch) {
if (ch == ‘A‘)
return 0;
if (ch == ‘C‘)
return 1;
if (ch == ‘G‘)
return 2;
return 3;
}
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] |= (1<<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]];
}
}
}
时间: 2024-10-13 04:40:54