4.7字符串上的动态规划
挑战程序竞赛系列（65）：4.7字符串上的动态规划（3）
题意：
基因工程：给定m个子串，求构造长n的母串的方案数。母串中每个字符都至少来自一个子串。
中文的解释有点含糊，不如看原文公式：
More formally: denote by |w| the length of w, let symbols of w be numbered from 1 to |w|. Then for each position i in w there exist pair of indices l, r (1 ≤ l ≤ i ≤ r ≤ |w|) such that the substring w[l … r] equals one of the elements s1, s2, …, sm of the collection.
所以说w中的每个字符，都能找到一个左边界和右边界属于某个子串即可。
思路：
dp[i][j] 在状态i下，后缀未能匹配的长度为j的方案数
所以，我们求的是各种状态下dp[i][0]之和
这里省去了阶段，因为下一阶段总由上一阶段生成，没必要重复记录。
代码很巧妙，记录了后缀的最大长度，解决了重叠问题。
如果后缀（状态）中不存在子串，最大长度为0，而我们知道一个新串一定从0开始构建的。
所以，对于后缀中最大长度为0的这些状态一定是转移的中间态，而一旦在转移过程中，状态的最大长度非零。
说明当中存在了子串，那么既然能够抵达该状态，长度为newNeed的新串一定属于该状态的某个最大子串中。代码如下：
import java.io.BufferedReader;
import java.io.File;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Map;
import java.util.Set;
import java.util.StringTokenizer;
public class Main{
String INPUT = "./data/judge/201709/C086C.txt";
public static void main(String[] args) throws IOException {
new Main().run();
}
static final int MOD = 1000000009;
int solve(int n, String[] p) {
int m = p.length;
//求出每个模式的后缀
Set<String> pfx = new HashSet<>();
pfx.add("");
int maxLength = 1;
for (int i = 0; i < m; ++i) {
maxLength = Math.max(maxLength, p[i].length());
for (int j = 1; j <= p[i].length(); ++j) {
pfx.add(p[i].substring(0, j));
}
}
int state = pfx.size();
Map<String, Integer> mem = new HashMap<>();
int idx = 0;
for (String pf : pfx) {
mem.put(pf, idx++);
}
//更新每个状态对应的最大子串长度
int[] finish = new int[state];
for (String pf : pfx) {
for (String pp : p) {
if (pf.endsWith(pp)) {
finish[mem.get(pf)] = Math.max(finish[mem.get(pf)], pp.length());
}
}
}
//状态转移
int[][] next = new int[state][4];
for (String pf : pfx) {
for (int ch = 0; ch < 4; ++ch) {
String nxt = pf + "ATCG".charAt(ch);
while (!mem.containsKey(nxt)) {
nxt = nxt.substring(1);
}
next[mem.get(pf)][ch] = mem.get(nxt);
}
}
int[][] dp = new int[state + 16][maxLength + 16];
dp[mem.get("")][0] = 1;
for (int i = 1; i <= n; ++i) {
dp = oneStep(dp, maxLength, state, next, finish);
}
int ans = 0;
for (int i = 0; i < state; ++i) {
ans += dp[i][0];
if (ans >= MOD) ans -= MOD;
}
return ans;
}
int[][] oneStep(int[][] cnt, int maxLength, int state, int[][] next, int[] finish) {
int[][] newDp = new int[state + 16][maxLength + 16];
for (int oldState = 0; oldState < state; ++oldState) {
for (int oldNeed = 0; oldNeed < maxLength; ++oldNeed) {
for (int ch = 0; ch < 4; ++ch) {
int newState = next[oldState][ch];
int newNeed = oldNeed + 1;
if (newNeed <= finish[newState]) newNeed = 0;
if (newNeed >= maxLength) continue;
newDp[newState][newNeed] += cnt[oldState][oldNeed];
if (newDp[newState][newNeed] >= MOD) newDp[newState][newNeed] -= MOD;
}
}
}
return newDp;
}
void read() {
int n = ni();
int m = ni();
String[] p = new String[m];
for (int i = 0; i < m; ++i) {
p[i] = ns();
}
out.println(solve(n, p));
}
FastScanner in;
PrintWriter out;
void run() throws IOException {
boolean oj;
try {
oj = ! System.getProperty("user.dir").equals("F:\\java_workspace\\leetcode");
} catch (Exception e) {
oj = System.getProperty("ONLINE_JUDGE") != null;
}
InputStream is = oj ? System.in : new FileInputStream(new File(INPUT));
in = new FastScanner(is);
out = new PrintWriter(System.out);
long s = System.currentTimeMillis();
read();
out.flush();
if (!oj){
System.out.println("[" + (System.currentTimeMillis() - s) + "ms]");
}
}
public boolean more(){
return in.hasNext();
}
public int ni(){
return in.nextInt();
}
public long nl(){
return in.nextLong();
}
public double nd(){
return in.nextDouble();
}
public String ns(){
return in.nextString();
}
public char nc(){
return in.nextChar();
}
class FastScanner {
BufferedReader br;
StringTokenizer st;
boolean hasNext;
public FastScanner(InputStream is) throws IOException {
br = new BufferedReader(new InputStreamReader(is));
hasNext = true;
}
public String nextToken() {
while (st == null || !st.hasMoreTokens()) {
try {
st = new StringTokenizer(br.readLine());
} catch (Exception e) {
hasNext = false;
return "##";
}
}
return st.nextToken();
}
String next = null;
public boolean hasNext(){
next = nextToken();
return hasNext;
}
public int nextInt() {
if (next == null){
hasNext();
}
String more = next;
next = null;
return Integer.parseInt(more);
}
public long nextLong() {
if (next == null){
hasNext();
}
String more = next;
next = null;
return Long.parseLong(more);
}
public double nextDouble() {
if (next == null){
hasNext();
}
String more = next;
next = null;
return Double.parseDouble(more);
}
public String nextString(){
if (next == null){
hasNext();
}
String more = next;
next = null;
return more;
}
public char nextChar(){
if (next == null){
hasNext();
}
String more = next;
next = null;
return more.charAt(0);
}
}
static class ArrayUtils {
public static void fill(int[][] f, int value) {
for (int i = 0; i < f.length; ++i) {
Arrays.fill(f[i], value);
}
}
public static void fill(int[][][] f, int value) {
for (int i = 0; i < f.length; ++i) {
fill(f[i], value);
}
}
public static void fill(int[][][][] f, int value) {
for (int i = 0; i < f.length; ++i) {
fill(f[i], value);
}
}
}
}
这里写图片描述
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原始发表：2017年09月14日，如有侵权请联系 cloudcommunity@tencent.com 删除
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