poj1639解题报告_SA

The answer is given in the 146th episode of Star Trek - The Next Generation, titled The Chase. It turns out that in the vast majority of the quadrant's life forms ended up with a large fragment of common DNA.
Given the DNA sequences of several life forms represented as strings of letters, you are to find the longest substring that is shared by more than half of them.
}
int checkok(int X,bool out){ int i,j,tmp; memset(cnt,0,sizeof(cnt)); for (i=j=0;i<L;i++){ if (lcp[i]<X){ tmp=i-j+1; if (tmp*2>N&&morethanhalf(j,i)){ if (out) printstring(s+SA[j],X); else return 1; } j=i+1; } } return 0;
并输出所有符合条件的串
解法：
此题本质上是要统计每个子串出现的次数，只需将其所有子串排序后扫描即可。即后 缀数组。
后缀数组 SA 是一个一维数组，它保存 1..n 的某个排列 SA[1],SA[2],...SA[n]，并
且保证 Suffix(SA[i])<Suffix(SA[i+1]),1≤i<n。也就是将 S 的 n 个后缀从小到大进行排 序之后把排好序的后缀的开头位置顺次放入 SA 中。
if (s[SA[i]]!=s[SA[i-1]]) Rank[SA[i]]=Rank[SA[i-1]]+1; else Rank[SA[i]]=Rank[SA[i-1]]; } for (k=1;k<L&&Rank[SA[L-1]]<L-1;k*=2){ for (i=0;i<L;i++) count[Rank[SA[i]]]=i+1; for (i=L-1;i>=0;i--) if (SA[i]>=k)
代码：
#include <stdio.h> #include <string.h> #define maxn 128 #define maxl 1024 #define maxsize 110000 int s[maxsize], id[maxsize]; char t[maxl]; int N,L; /* For SA */ int count[maxsize]; int mem[3][maxsize],lcp[maxsize]; int *SA,*nSA,*Rank,*nRank; int cnt[maxn]; void suffix_array(){
Output
For each test case, output the longest string or strings shared by more than half of the life forms. If there are many, output all of them in alphabetical order. If there is no solution with at least one letter, output "?". Leave an empty line between test cases.
else nRank[SA[i]]=nRank[SA[i-1]];
} nSA=Rank; Rank=nRank; } for (i=0,k=0;i<L;i++) if (Rank[i]==L-1) lcp[Rank[i]]=k=0; else{
if (k>0) k--; j=SA[Rank[i]+1]; for (;s[i+k]==s[j+k];k++); lcp[Rank[i]]=k; } }
较好的方法是倍增算法，基本思想是把 n 个后缀按照 k-前缀排序，将排序后的后缀的 开头位置顺次放入数组 SA_k 中，称之为 k-后缀数组，其逆函数 Rank_k 称为 k-名次数组。
求出 Rank_k 后，可以采用基数排序在 O(n)时间内求出 Rank_2k，重复此过程至多 log(n)次，就可以从 Rank_1 推出最终的名次数组 Rank。
nSA[--count[Rank[SA[i]-k]]]=SA[i]-k; for (i=L-k;i<L;i++) nSA[--count[Rank[i]]]=i; nRank=SA; SA=nSA; for (nRank[SA[0]]=0,i=1;i<L;i++){
if (Rank[SA[i]]!=Rank[SA[i-1]]||Rank[SA[i]+k]!=Rank[SA[i-1]+k]) nRank[SA[i]]=nRank[SA[i-1]]+1;
Input
Standard input contains several test cases. Each test case begins with 1 ≤ n ≤ 100, the number of life forms. n lines follow; each contains a string of lower case letters representing the DNA sequence of a life form. Each DNA sequence contains at least one and not more than 1000 letters. A line containing 0 follows the last test case.
int i,j,k; SA=mem[0]; nSA=mem[1]; Rank=mem[2]; memset(count,0,sizeof(int)*300); for (i=0;i<L;i++) count[s[i]]++; for (i=1;i<300;i++) count[i]+=count[i-1]; for (i=0;i<L;i++) SA[--count[s[i]]]=i; for (Rank[SA[0]]=0,i=1;i<L;i++){
Sample Input
3 abcdefg bcdefgh cdefghi 3 xxx yyy zzz 0
Sample Output
bcdefg cdefgh
?
Source
Waterloo Local Contest, 2006.9.30
题目大意
给 n(n<=100)个长度不超过 m(m<=1000)的字符串，求超过一半串的最长公共子串，
void printstring(int *ss,int sl){ int i; for (i=0;i<sl;i++) printf("%c",ss[i]); printf("\n");
}
bool morethanhalf(int begin,int end){ int i,rs; memset(cnt,0,sizeof(cnt)); for (i=begin,rs=0;i<=end;i++){ if (cnt[id[SA[i]]]==0){ cnt[id[SA[i]]]++; rs++; } } return rs*2>N;
名次数组 Rank 是 SA 的逆函数，也就是说若 SA[i]=j，则 Rank[j]=i，不难看出
Rank[i]保存的是 Suffix(i)在所有后缀中从小到大排列的“名次”。 可以像标程那样对 S 的所有后缀用 qsort()暴力排序，这种做法虽然很看起来土，理
论上最坏情况下的复杂度为高达 L^2*log(L)，但常数项实际上很小，只要稍加优化对本 题的数据规模仍然可以很容易 AC。
题目原文
Life Forms
Time 5000MS
Limit: Memory 65536K
Limit:
Total Submissions: 887
Accepted: 241
Description
You may have wondered why most extraterrestrial life forms resemble humans, differing by superficial traits such as height, colour, wrinkles, ears, eyebrows and the like. A few bear no human resemblance; these typically have geometric or amorphous shapes like cubes, oil slicks or clouds of dust.
回到原问题，首先，将所有字符串连接起来，为了避免前后影响，需要在每个字符串 的末尾加入一个互不相同的不会在原文中出现的字符，例如 0x80+ID，来分隔不同的字 符串。求此长串的后缀数组，并求得相邻串的最长公共前缀，并由此二分求出所有连续 n/2 区间的最长公共前缀的长度即可。
时间复杂度：