zoukankan      html  css  js  c++  java
  • POJ3294:Life Forms(后缀数组)

    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


    给定n个字符串,求出如今不小于k/2个字符串中的最长子串。 
    将n个字符串连起来,中间用不同样的且没有出如今字符串中的字符隔开, 求后缀数组。
    然后二分答案,将后缀分成若干组,推断每组的后缀是否出如今不小于k个的原串中。

    这个做法的时间复杂度为O(nlogn)。



    #include <iostream>
    #include <stdio.h>
    #include <string.h>
    #include <stack>
    #include <queue>
    #include <map>
    #include <set>
    #include <vector>
    #include <math.h>
    #include <bitset>
    #include <algorithm>
    #include <climits>
    using namespace std;
    
    #define LS 2*i
    #define RS 2*i+1
    #define UP(i,x,y) for(i=x;i<=y;i++)
    #define DOWN(i,x,y) for(i=x;i>=y;i--)
    #define MEM(a,x) memset(a,x,sizeof(a))
    #define W(a) while(a)
    #define gcd(a,b) __gcd(a,b)
    #define LL long long
    #define N 1000005
    #define MOD 1000000007
    #define INF 0x3f3f3f3f
    #define EXP 1e-8
    int wa[N],wb[N],wsf[N],wv[N],sa[N];
    int rank[N],height[N],s[N],a[N];
    //sa:字典序中排第i位的起始位置在str中第sa[i]
    //rank:就是str第i个位置的后缀是在字典序排第几
    //height:字典序排i和i-1的后缀的最长公共前缀
    int cmp(int *r,int a,int b,int k)
    {
        return r[a]==r[b]&&r[a+k]==r[b+k];
    }
    void getsa(int *r,int *sa,int n,int m)//n要包括末尾加入的0
    {
        int i,j,p,*x=wa,*y=wb,*t;
        for(i=0; i<m; i++)  wsf[i]=0;
        for(i=0; i<n; i++)  wsf[x[i]=r[i]]++;
        for(i=1; i<m; i++)  wsf[i]+=wsf[i-1];
        for(i=n-1; i>=0; i--)  sa[--wsf[x[i]]]=i;
        p=1;
        j=1;
        for(; p<n; j*=2,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<n; i++)  wv[i]=x[y[i]];
            for(i=0; i<m; i++)  wsf[i]=0;
            for(i=0; i<n; i++)  wsf[wv[i]]++;
            for(i=1; i<m; i++)  wsf[i]+=wsf[i-1];
            for(i=n-1; i>=0; i--)  sa[--wsf[wv[i]]]=y[i];
            t=x;
            x=y;
            y=t;
            x[sa[0]]=0;
            for(p=1,i=1; i<n; i++)
                x[sa[i]]=cmp(y,sa[i-1],sa[i],j)? p-1:p++;
        }
    }
    void getheight(int *r,int n)//n不保存最后的0
    {
        int i,j,k=0;
        for(i=1; i<=n; i++)  rank[sa[i]]=i;
        for(i=0; i<n; i++)
        {
            if(k)
                k--;
            else
                k=0;
            j=sa[rank[i]-1];
            while(r[i+k]==r[j+k])
                k++;
            height[rank[i]]=k;
        }
    }
    
    char str[N];
    int len[105],size,ans[N];
    bool vis[105];
    
    int check(int mid,int n,int k)
    {
        int i,j;
        int size = 0,cnt = 0;
        MEM(vis,false);
        for(i = 1; i<=n; i++)
        {
            if(height[i]>=mid)
            {
                for(j = 1; j<=k; j++)
                {
                    //把sa[i-1]或sa[i]所在的字符串给标记。相同的串不反复累加
                    if(sa[i]>len[j-1]&&sa[i]<len[j]) cnt+=(vis[j]?0:1),vis[j]=true;
                    if(sa[i-1]>len[j-1]&&sa[i-1]<len[j]) cnt+=(vis[j]?0:1),vis[j]=true;
                }
            }
            else
            {
                if(cnt>k/2) ans[++size] = sa[i-1];
                cnt = 0;
                MEM(vis,false);
            }
        }
        if(cnt>k/2) ans[++size] = sa[n];
        if(size)
        {
            ans[0] = size;
            return 1;
        }
        return 0;
    }
    
    int main()
    {
        int n,k,i,j,flag = 0;
        while(~scanf("%d",&k),k)
        {
            n = 0;
            size = 0;
            for(i = 1; i<=k; i++)
            {
                scanf("%s",str+n);
                for(; str[n]!=''; n++)
                    s[n] = str[n];
                s[n] = '#'+i;
                len[++size] = n;
                n++;
            }
            s[n-1] = 0;
            getsa(s,sa,n,255);
            getheight(s,n-1);
            int l=1,r=n,mid;
            while(l<=r)
            {
                mid = (l+r)/2;
                if(check(mid,n,k)) l = mid+1;
                else r = mid-1;
            }
            if(flag)
                puts("");
            flag = 1;
            if(l==1)
                puts("?

    "); else { for(i = 1; i<=ans[0]; i++) { for(j = ans[i]; j<ans[i]+l-1; j++) printf("%c",s[j]); puts(""); } } } return 0; }



  • 相关阅读:
    IPv4地址被用光,IPv6将接手
    杀猪盘
    大家都应该看看这个贴子,会让你心明眼亮。 注意到这些变化了吗?中国正在发生的100个变化,越往后读越震惊!
    区块链在中国怎么练?
    区块链到底是什么样的技术呢?
    2019感恩节
    人工智能、大数据、物联网、区块链,四大新科技PK,你更看好谁?
    vue遇见的问题(2)---imported multiple times(转载)
    drf-序列化器的理解
    Django rest_framework序列化many=True参数解释
  • 原文地址:https://www.cnblogs.com/yangykaifa/p/6789264.html
Copyright © 2011-2022 走看看