Life Forms
|
Time Limit: 5000MS |
|
Memory Limit: 65536K |
|
Total Submissions: 11652 |
|
Accepted: 3228 |
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.
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.
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.
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.
Sample Input
3
abcdefg
bcdefgh
cdefghi
3
xxx
yyy
zzz
0
Sample Output
bcdefg
cdefgh
?
Source
Waterloo Local Contest, 2006.9.30
【思路】
多个字符串求出现超过R次的最长公共子串。
二分+划分height，判定一个组中是否包含不小于R个不同字符串的后缀。
需要注意的有：
1) c[]尽量开大，字符范围为“偏移”之后的范围。
2) 用kase作为标记节省了每次开始新段需要清零的时间。
3) 因为height是sa[i]与sa[i-1]的关系，所以无论是在can的开始还是在新段开始都需要初始为一个串的情况。
【代码】
【代码】
1 #include<cstdio> 2 #include<cstring> 3 #include<vector> 4 #include<iostream> 5 #define FOR(a,b,c) for(int a=(b);a<=(c);a++) 6 using namespace std; 7 8 const int maxn = 200000+10; 9 10 int s[maxn]; 11 int sa[maxn],c[maxn],t[maxn],t2[maxn]; 12 13 void build_sa(int m,int n) { 14 int i,*x=t,*y=t2; 15 for(i=0;i<m;i++) c[i]=0; 16 for(i=0;i<n;i++) c[x[i]=s[i]]++; 17 for(i=1;i<m;i++) c[i]+=c[i-1]; 18 for(i=n-1;i>=0;i--) sa[--c[x[i]]]=i; 19 20 for(int k=1;k<=n;k<<=1) { 21 int p=0; 22 for(i=n-k;i<n;i++) y[p++]=i; 23 for(i=0;i<n;i++) if(sa[i]>=k) y[p++]=sa[i]-k; 24 25 for(i=0;i<m;i++) c[i]=0; 26 for(i=0;i<n;i++) c[x[y[i]]]++; 27 for(i=0;i<m;i++) c[i]+=c[i-1]; 28 for(i=n-1;i>=0;i--) sa[--c[x[y[i]]]]=y[i]; 29 30 swap(x,y); 31 p=1; x[sa[0]]=0; 32 for(i=1;i<n;i++) 33 x[sa[i]]=y[sa[i]]==y[sa[i-1]] && y[sa[i]+k]==y[sa[i-1]+k]?p-1:p++; 34 if(p>=n) break; 35 m=p; 36 } 37 } 38 int rank[maxn],height[maxn]; 39 void getHeight(int n) { 40 int i,j,k=0; 41 for(i=0;i<=n;i++) rank[sa[i]]=i; 42 for(i=0;i<n;i++) { 43 if(k) k--; 44 j=sa[rank[i]-1]; 45 while(s[j+k]==s[i+k]) k++; 46 height[rank[i]]=k; 47 } 48 } 49 50 int T; 51 char a[maxn]; 52 53 int f[200],kase; 54 vector<int> st; 55 int can(int limit,int n,int len) { 56 int cnt=1,ok=0; 57 st.clear(); 58 f[sa[1]/len]=kase; 59 for(int i=2;i<=n;i++) { 60 if(height[i]<limit) { 61 cnt=1; 62 f[sa[i]/len]=++kase; //检查每一个组中 63 } 64 else { 65 if(f[sa[i]/len]!=kase) { 66 f[sa[i]/len]=kase; 67 if(cnt>=0) cnt++; 68 if(cnt>T/2) { 69 ok=1; 70 st.push_back(sa[i]); 71 cnt=-1; 72 } 73 } 74 } 75 } 76 return ok; 77 } 78 void init() { 79 kase=1; 80 memset(sa,0,sizeof(sa)); 81 memset(f,0,sizeof(f)); 82 } 83 int main() { 84 //freopen("in.in","r",stdin); 85 //freopen("out.out","w",stdout); 86 while(scanf("%d",&T)==1 && T) { 87 init(); 88 int len,n=0; 89 for(int i=0;i<T;i++) { 90 scanf("%s",&a); 91 len=strlen(a); 92 for(int j=0;j<len;j++) s[n++]=a[j]+100; 93 s[n++]=i+1; 94 } 95 n--; 96 s[n]=0; 97 98 build_sa(250,n+1); 99 getHeight(n); 100 101 int L=0,R=len+1; 102 while(L<R) { 103 int M=L+(R-L+1)/2; 104 if(can(M,n,len+1)) L=M; 105 else R=M-1; 106 } 107 can(L,n,len+1); //再调用一次求出st 108 if(L==0) printf("?\n"); 109 else { 110 for(int i=0;i<st.size();i++) { 111 for(int j=st[i];(j-st[i]+1)<=L;j++) 112 printf("%c",s[j]-100); 113 putchar(‘\n‘); 114 } 115 } 116 putchar(‘\n‘); 117 } 118 return 0; 119 }
原文：http://www.cnblogs.com/lidaxin/p/5004442.html