[UVA][Trie] 760 - DNA Sequencing＠Morris' Blog｜PChome Online 人新台

Morris' Blog 我在摸索我存在的意、生命的意、涵及值。有真正神手般的害，套用模式是大部分的情。那害的我存在的意何，托出神的存在？有候不能怪人不我，我已始省思，什我有人的原因，而是全要靠人。有候我我不同，那是因我做不到很多人能到的事情。我不到、玩不起、得不明、理解得不多，我缺少的文能力，就是一切一切能的判定。此，我不得能做些人有助的事情。我知道我自己是不服的性格，但在我不得不。我有能力，因此我必。我到底世界作什？充不定性，而我是法出拔萃，想一般人似乎也回不了了。越大的挑需要越力的支持，而我的支持是什？最近人，但都是失的局，人能接受，不需要言，不需要通，一切所想都表不出。
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2013-02-26 08:26:42| 人1,442| 回0 | 上一篇 | 下一篇

[UVA][Trie] 760 - DNA Sequencing

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DNA Sequencing

A DNA molecule consists of two strands that wrap around each other to resemble a twisted ladder whose sides, made of sugar and phosphate molecules, are connected by rungs of nitrogen-containing chemicals called bases. Each strand is a linear arrangement of repeating similar units called nucleotides, which are each composed of one sugar, one phosphate, and a nitrogenous base. Four different bases are present in DNA: adenine (A), thymine (T), cytosine (C), and guanine (G). The particular order of the bases arranged along the sugar-phosphate backbone is called the DNA sequence; the sequence specifies the exact genetic instructions required to create a particular organism with its own unique traits.

Geneticists often compare DNA strands and are interested in finding the longest common base sequence in the two strands. Note that these strands can be represented as strings consisting of the letters a, t, c and g. So, the longest common sequence in the two strands atgc and tga is tg. It is entirely possible that two different common sequences exist that are the same length and are the longest possible common sequences. For example in the strands atgc and gctg, the longest common sequences are gc and tg.

Input and Output

Write a program that accepts as input two strings representing DNA strands, and prints as output the longest common sequence(s) in lexicographical order.

If there isn't any common sequence between the two strings, just print: ``No common sequence."

If there are more than one test cases, it must be a blank line between two consecutive, both in input and output files.

The strings are at most 300 characters-long.

Sample Input

atgc tga atgc gctg

Sample Output

tg gc tg

原本要用 SA 去做的，但很小，length < 300。
用 O(n*n) 建，一用 O(n*n) 去 match 棵。
然後出即可。

很妙的是，我一始就 RE 了，但是 UVa 上是 WA，我一直抓不到 BUG。


#include <stdio.h>
#include <string.h>
struct TrieNd{
 int LK[26];
 int ac;
} ND[100000];
int TrieSize = 0;
void insTrie(char s[]) {
 static int i, idx, v;
 for(i = 0, idx = 0; s[i]; i++) {
 v = s[i]-'a';
 if(!ND[idx].LK[v]) {
 TrieSize++;
 memset(&ND[TrieSize], 0, sizeof(ND[0]));
 ND[idx].LK[v] = TrieSize;
 }
 idx = ND[idx].LK[v];
 }
}
void matchTrie(char s[]) {
 static int i, idx, v;
 for(i = 0, idx = 0; s[i]; i++) {
 v = s[i]-'a';
 ND[idx].ac = 1;
 if(ND[idx].LK[v])
 idx = ND[idx].LK[v];
 else
 break;
 }
 ND[idx].ac = 1;
}
int mxlen = -1;
void dfs(int nd, int dep) {
 int i;
 for(i = 0; i < 26; i++) {
 if(ND[nd].LK[i] && ND[ND[nd].LK[i]].ac) {
 if(dep+1 > mxlen)
 mxlen = dep+1;
 dfs(ND[nd].LK[i], dep+1);
 }
 }
}
char path[512];
void dfs2(int nd, int dep) {
 int i;
 for(i = 0; i < 26; i++) {
 if(ND[nd].LK[i] && ND[ND[nd].LK[i]].ac) {
 path[dep] = i+'a';
 if(dep+1 == mxlen) {
 path[dep+1] = '\0';
 puts(path);
 }
 dfs2(ND[nd].LK[i], dep+1);
 }
 }
}
int main() {
 char s1[512], s2[512];
 int i, first = 0;
 while(gets(s1)){
 gets(s2);
 if(first) puts("");
 first = 1;
 memset(&ND[0], 0, sizeof(ND[0]));
 TrieSize = 0;
 for(i = 0; s1[i]; i++)
 insTrie(s1+i);
 for(i = 0; s2[i]; i++)
 matchTrie(s2+i);
 mxlen = -1;
 dfs(0, 0);
 if(mxlen > 0)
 dfs2(0, 0);
 else
 puts("No common sequence.");
 gets(s1);
 }
 return 0;
}