poj3294 --Life Forms
| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 12483 | Accepted: 3501 |
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 ?
瘠薄
1 #include<cstdio> 2 #include<cstring> 3 #include<cmath> 4 #include<string> 5 #include<algorithm> 6 #include<iostream> 7 #define maxn 200005 8 int ws[maxn],wa[maxn],sa[maxn],num[maxn],n,wv[maxn],rank[maxn]; 9 int h[maxn],wb[maxn],sum[maxn],m; 10 char str[105][1005]; 11 bool cmp(int *r,int a,int b,int l){ 12 return r[a]==r[b]&&r[a+l]==r[b+l]; 13 } 14 39 void da(int *r,int *sa,int n,int m){ 40 int *t,*x=wa,*y=wb,i,j,p; 41 for (i=0;i<m;i++) ws[i]=0; 42 for (i=0;i<n;i++) x[i]=r[i]; 43 for (i=0;i<n;i++) ws[x[i]]++; 44 for (i=1;i<m;i++) ws[i]+=ws[i-1]; 45 for (i=n-1;i>=0;i--) sa[--ws[x[i]]]=i; 46 for (j=1,p=1;p<n;j*=2,m=p){ 47 for (p=0,i=n-j;i<n;i++) y[p++]=i; 48 for (i=0;i<n;i++) if (sa[i]-j>=0) y[p++]=sa[i]-j; 49 for (i=0;i<m;i++) ws[i]=0; 50 for (i=0;i<n;i++) wv[i]=x[y[i]]; 51 for (i=0;i<n;i++) ws[wv[i]]++; 52 for (i=1;i<m;i++) ws[i]+=ws[i-1]; 53 for (i=n-1;i>=0;i--) sa[--ws[wv[i]]]=y[i]; 54 for (t=x,x=y,y=t,i=1,p=1,x[sa[0]]=0;i<n;i++) 55 x[sa[i]]=cmp(y,sa[i],sa[i-1],j)?p-1:p++; 56 } 57 } 58 void cal(int *r,int n){ 59 int i,j,k=0; 60 for (int i=1;i<=n;i++) rank[sa[i]]=i; 61 for (int i=0;i<n;h[rank[i++]]=k) 62 for (k?k--:0,j=sa[rank[i]-1];r[i+k]==r[j+k];k++); 63 } 64 int getid(int k){ 65 int l=0,r=n-1,mid; 66 while (l<r){ 67 mid=(l+r)/2; 68 if (sum[mid]<k) { 69 l=mid+1; 70 } 71 else 72 r=mid; 73 } 74 return l; 75 } 76 77 bool check(int len,int out=0){ 78 int i=n+1,j,k,id,cnt; 79 bool f[maxn]; 80 while (1){ 81 while (i<=m&&h[i]<len) i++; 82 if (i>m) break; 83 memset(f,0,sizeof f); 84 id=getid(sa[i-1]); 85 f[id]=true; 86 cnt=1; 87 while (i<=m&&h[i]>=len){ 88 id=getid(sa[i]); 89 if (!f[id]){ 90 f[id]=true; 91 cnt++; 92 } 93 i++; 94 } 95 if (out==0){ 96 if (2*cnt>n) return true; 97 } 98 else 99 if (2*cnt>n){ 100 for (k=sa[i-1],j=0;j<len;k++,j++){ 101 printf("%c",num[k]+'a'-100); 102 } 103 printf("\n"); 104 } 105 106 } 107 return false; 108 } 144 int main(){ 145 freopen("tx.in","r",stdin); 146 int i,j,k; 147 while (scanf("%d",&n)&&n!=0){ 148 scanf("%s",str[0]); 149 if (n==1){ 150 printf("%s\n\n",str[0]); 151 continue; 152 } 153 sum[0]=strlen(str[0]); 154 for (i=1;i<n;i++){ 155 scanf("%s",str[i]); 156 sum[i]=sum[i-1]+strlen(str[i])+1; 157 } 158 for (k=i=0;i<n;i++){ 159 for (j=0;j<strlen(str[i]);j++){ 160 num[k++]=str[i][j]-'a'+100; 161 } 162 num[k++]=i; 163 } 164 m=k-1; 165 da(num,sa,m+1,130); 166 cal(num,m); 167 int l=0,r=m,mid; 168 while (l<r){ 169 mid=(l+r+1)/2; 170 if (check(mid)) l=mid; 171 else r=mid-1; 172 } 173 if (l==0) printf("?\n\n"); 174 else { 175 check(l,1); 176 printf("\n"); 177 } 178 } 179 }
后缀数组
