UVA 11107 Life Forms 后缀数组

白书上的例题，贴的模板。。。

Life Forms

Time Limit: 6666MS Memory Limit: Unknown 64bit IO Format: %lld & %llu

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Description

Problem C: Life Forms

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.

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.

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

3abcdefgbcdefghcdefghi3xxxyyyzzz0

Output for Sample Input

bcdefgcdefgh?

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Gordon V. Cormack

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#include <cstdlib>#include <cctype>#include <cstring>#include <cstdio>#include <cmath>#include <algorithm>#include <vector>#include <string>#include <iostream>#include <sstream>#include <map>#include <set>#include <queue>#include <stack>#include <fstream>#include <numeric>#include <iomanip>#include <bitset>#include <list>#include <stdexcept>#include <functional>#include <utility>#include <ctime>using namespace std;#define PB push_back#define MP make_pair#define REP(i,n) for(int i=0;i<(n);++i)#define FOR(i,l,h) for(int i=(l);i<=(h);++i)#define DWN(i,h,l) for(int i=(h);i>=(l);--i)#define CLR(vis) memset(vis,0,sizeof(vis))#define MST(vis,pos) memset(vis,pos,sizeof(vis))#define MAX3(a,b,c) max(a,max(b,c))#define MAX4(a,b,c,d) max(max(a,b),max(c,d))#define MIN3(a,b,c) min(a,min(b,c))#define MIN4(a,b,c,d) min(min(a,b),min(c,d))#define PI acos(-1.0)#define INF 0x7FFFFFFF#define LINF 1000000000000000000LL#define eps 1e-8typedef long long ll;const int maxn=100000+1000;int r[maxn] ;int sa[maxn], height[maxn] ,rank[maxn] , id[maxn] ;int wa[maxn] ,wb[maxn], ky[maxn], cn[maxn] ;inline bool cmp(int *r ,int a,int b,int l){    return r[a]==r[b] && r[a+l] == r[b+l] ;}void build_sa(int *r, int *sa ,int n , int m){    int i, k , p , *x=wa ,*y = wb;    for(i=0; i<m; i++) cn[i] = 0;    for(i=0; i<n; i++) cn[ x[i] = r[i] ] ++ ;    for(i=1; i<m; i++) cn[i] += cn[i-1] ;    for(i=n-1; i>=0; i--) sa[ --cn[ x[i] ] ] = i ;    for(k=1,p=1; p<n; k<<=1 , m=p) {        for(p=0,i=n-k; i<n; i++) y[p++] = i ;        for(i=0; i<n; i++) if(sa[i] >=k ) y[p++] = sa[i] - k ;        for(i=0; i<n; i++) ky[i] = x[ y[i] ] ;        for(i=0; i<m; i++) cn[i] = 0;        for(i=0; i<n; i++) cn[ky[i]]++ ;        for(i=1; i<m; i++) cn[i] += cn[i-1] ;        for(i=n-1; i>=0; i--) sa[ --cn[ky[i]] ] = y[i] ;        for(swap(x ,y),p=1,x[sa[0]]=0,i=1; i<n; i++)        x[sa[i]] = cmp(y,sa[i-1],sa[i],k) ? p-1 : p++ ;    }    return ;}void get_height(int *s, int n) {    int i, j , k=0;    for(i=0; i<n; i++) rank[sa[i]] = i ;    for(i=0; i<n-1; i++) {        if(k) k-- ;        j = sa[rank[i]-1] ;        while(s[i+k] == s[j+k]) k++ ;        height[rank[i]] = k;    }}bool check(int n ,int m, int mid){    set<int>vis ;    vis.insert(id[sa[1]]) ;    for(int i=2;i<n;i++)    {        while(i<n && height[i] >= mid)          vis.insert(id[sa[i]]),i++ ;        if(vis.size()*2>m)           return true ;        vis.clear();        vis.insert(id[sa[i]]) ;    }    return false ;}void print(int n,int m,int p){    set<int>vis ;    vis.insert(id[sa[1]]);    for(int i=2; i<n; i++) {        while(i<n && height[i]>= p)            vis.insert(id[sa[i]]),i++ ;        if(vis.size()*2>m)        {            int a=sa[i-1] ;            for(int j=0;j<p;j++)               putchar(r[a+j]) ;            puts("") ;        }        vis.clear() ;        vis.insert(id[sa[i]]) ;    }}int main(){    int m,flag=1;    while(scanf("%d",&m)==1 && m)    {        if(!flag) puts("");        else     flag=0;        char str[1111];        int n=0;        FOR(i,1,m)        {            CLR(str);            scanf("%s",str);            for(int j=0; str[j]; ++j)            {                id[n]=i;                r[n++]=str[j];            }            id[n]=i;            r[n++]='z'+i;        }        if(m==1)        {            puts(str);            continue;        }        id[n]=n+1;        r[n++]=0;        build_sa(r,sa,n,'z'+m+2);        get_height(r,n);        if(!check(n,m,1))        {            puts("?");            continue;        }        int L=1,R=1000,mid;        while(L<R)        {            mid=L+(R-L+1)/2;            if(check(n,m,mid))                L=mid ;            else                R=mid-1;        }        print(n,m,L);    }    return 0;}