uva 11107 - Life Forms

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?

这道题有点坑，首先是我无法理解more than a half。最后能AC的程序是这样算的，如果只有1个字符串，那么more than a half算1。

用（1,1）表示，然后是（2,2），（3,2），（4,3）算是看出规律了（n，n/2+1）。好吧，就这么理解吧。

这题大白上有题解，这道题关键在于记录新字符串中每个字符对应的是哪个原串，用idx数组记录，再输入数据时就可以做完这事。

然后就是二分答案，t=1时，必须特别计算。

我是先找到答案maxlen的，然后再查一遍height数组，把每个分组里原串个数>=t/2+1的输出来。

总感觉代码还是有bug，数据水了点就AC了。

代码：

#include<cstdio>#include<cstring>#include<iostream>#define Maxn 1000200using namespace std;int r[Maxn],sa[Maxn],Rank[Maxn],height[Maxn];int wa[Maxn],wb[Maxn],rs[Maxn],wv[Maxn];char str[1010];int cmp(int *r,int a,int b,int l){    return r[a]==r[b]&&r[a+l]==r[b+l];}void da(int n,int m){    int i,j,p,*x=wa,*y=wb;    for(i=0;i<m;i++) rs[i]=0;    for(i=0;i<n;i++) rs[x[i]=r[i]]++;    for(i=1;i<m;i++) rs[i]+=rs[i-1];    for(i=n-1;i>=0;i--) sa[--rs[x[i]]]=i;    for(j=1,p=1;p<n;j<<=1,m=p){        for(p=0,i=n-j;i<n;i++) y[p++]=i;        for(i=0;i<n;i++) if(sa[i]>=j) y[p++]=sa[i]-j;        for(i=0;i<m;i++) rs[i]=0;        for(i=0;i<n;i++) rs[wv[i]=x[y[i]]]++;        for(i=1;i<m;i++) rs[i]+=rs[i-1];        for(i=n-1;i>=0;i--) sa[--rs[wv[i]]]=y[i];        swap(x,y);        for(p=1,x[sa[0]]=0,i=1;i<n;i++)            x[sa[i]]=cmp(y,sa[i-1],sa[i],j)?p-1:p++;    }}void calheight(int n){    int i,j,k=0;    for(int i=1;i<n;i++) Rank[sa[i]]=i;    for(int i=1;i<n;height[Rank[i++]]=k){if(k) k--;        for(j=sa[Rank[i]-1];r[i+k]==r[j+k];k++);    }}int idx[Maxn];int add(int x,int cur){    for(int i=0;str[i];i++){        r[cur]=str[i];        idx[cur++]=x;    }    r[cur++]=125+x;    return cur;}bool check(int n,int mid,int num){    int cnt=1,f[110];    memset(f,0,sizeof f);    f[idx[sa[1]]]=true;    for(int i=2;i<n;i++)        if(height[i]<mid) {cnt=1;memset(f,0,sizeof f);f[idx[sa[i]]]=true;}        else{            if(!f[idx[sa[i]]]) {cnt++;f[idx[sa[i]]]=true;}            if(cnt>=num) return true;        }    return false;}void print(int n,int len,int num){    int cnt=1,f[110];    memset(f,0,sizeof f);    f[idx[sa[1]]]=true;    bool flag=true;    for(int i=2;i<n;i++)        if(height[i]<len) {cnt=1;flag=true;memset(f,0,sizeof f);f[idx[sa[i]]]=true;}        else{            if(!f[idx[sa[i]]]) {cnt++;f[idx[sa[i]]]=true;}            if(cnt>=num&&flag){                for(int j=0;j<len;j++)                    printf("%c",r[sa[i]+j]);                puts("");                flag=false;            }        }}int main(){    int t,cas=0;    while(scanf("%d",&t),t){        if(cas++) puts("");        int cur=1;        for(int i=0;i<t;i++){            scanf("%s",str);            cur=add(i,cur);        }        if(t==1) {printf("%s\n",str);continue;}        r[0]=r[cur]=0;        da(cur,250);        calheight(cur);        int l=0,r=cur;        while(l<r){            int mid=l+r+1>>1;            if(check(cur,mid,(t>>1)+1)) l=mid;            else r=mid-1;        }        if(l==0) puts("?");        else print(cur,l,(t>>1)+1);    }return 0;}