POJ 3294 Life Forms [最长公共子串加强版 后缀数组 && 二分]

题目：http://poj.org/problem?id=3294

Life Forms

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 18549		Accepted: 5454

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

题意概括：

给出 N 个字符串，求其中出现次数超过 N/2 次的最长公共子串，如果有多种输出多种。

解题思路：

做法依然是二分答案长度，关键在于判断条件有两个：

①出现次数是否大于 N/2，这个通过height分组，统计一下即可。

②当前所枚举的子串不仅要求不能重叠，而且要满足来源于原本不同的字符串（因为合并了所有字符串，所以以原来字符串分区，判断两个子串要在不同区）

二分不重叠相同子串的加强版，网上很多版本都是暴力 O( n ) 判断子串是否来自不同串的，复杂度有点爆炸。

这道题复杂度的优化关键在于优化这个判断条件。

有个技巧：合并字符串时在中间加入分隔标志，后面通过 O(1) 标记即可判断是否满足区间要求。

输出子串的话，只要保存满足条件的 sa 即可。

AC code：

#include <set>
#include <map>
#include <cmath>
#include <vector>
#include <cstdio>
#include <cstring>
#include <string>
#include <iostream>
#include <algorithm>
#define INF 0x3f3f3f3f
#define LL long long
#define inc(i, j, k) for(int i = j; i <= k ; i++)
#define mem(i, j) memset(i, j, sizeof(i))
#define gcd(i, j) __gcd(i, j)
#define F(x) ((x)/3+((x)%3==1?0:tb))
#define G(x) ((x)<tb?(x)*3+1:((x)-tb)*3+2)
using namespace std;
const int MAXN = 3e5+10;
const int maxn = 3e5+10;
int r[MAXN];
int wa[MAXN], wb[MAXN], wv[MAXN], tmp[MAXN];
int sa[MAXN]; //index range 1~n value range 0~n-1
int cmp(int *r, int a, int b, int l)
{
    return r[a] == r[b] && r[a + l] == r[b + l];
}

void da(int *r, int *sa, int n, int m)
{
    int i, j, p, *x = wa, *y = wb, *ws = tmp;
    for (i = 0; i < m; i++) ws[i] = 0;
    for (i = 0; i < n; i++) ws[x[i] = r[i]]++;
    for (i = 1; i < m; i++) ws[i] += ws[i - 1];
    for (i = n - 1; i >= 0; i--) sa[--ws[x[i]]] = i;
    for (j = 1, p = 1; 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++) ws[i] = 0;
        for (i = 0; i < n; i++) ws[wv[i]]++;
        for (i = 1; i < m; i++) ws[i] += ws[i - 1];
        for (i = n - 1; i >= 0; i--) sa[--ws[wv[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], j) ? p - 1 : p++;
    }
}

int Rank[MAXN]; //index range 0~n-1 value range 1~n
int height[MAXN]; //index from 1   (height[1] = 0)
void calheight(int *r, int *sa, int n)
{
    int i, j, k = 0;
    for (i = 1; i <= n; ++i) Rank[sa[i]] = i;
    for (i = 0; i < n; height[Rank[i++]] = k)
        for (k ? k-- : 0, j = sa[Rank[i] - 1]; r[i + k] == r[j + k]; ++k);
    return;
}

int N;
string tp;
vector<int>ans_id;
int f[MAXN], kase;

bool check(int limit, int n, int len)
{
    bool flag = false;
    int cnt = 1;
    ans_id.clear();
    f[sa[1]/len] = kase;
    for(int i = 1; i <= n; i++){
        if(height[i] < limit){          //按height分组
                f[sa[i]/len] = ++kase;      //给区间标记上组的标号
                cnt = 1;
        }
        else{
            if(f[sa[i]/len] != kase){       //判断一组中是否有相同区间
                f[sa[i]/len] = kase;
                if(cnt>=0) cnt++;
                if(cnt > N/2){
                    flag = true;
                    ans_id.push_back(sa[i]);
                    cnt = -1;
                }
            }
        }
    }
    return flag;
}

int main()
{
    bool book = false;
    int ssize, n_len = 0, ans;
    while(~scanf("%d", &N) && N){
        n_len = 0;
        kase = 1;
        ans = 0;
        for(int i = 1; i <= N; i++){
            cin >> tp;
            ssize = tp.size();
            for(int k = 0; k < ssize; k++){
                r[n_len++] = tp[k]+100;
            }
            r[n_len++] = i;                 //作分隔标记
        }
        n_len--;
        r[n_len] = 0;

        da(r, sa, n_len+1, 277);
        calheight(r, sa, n_len);

        int L = 0, R = ssize+1, mid;
        while(L <= R){
            mid = (L+R)>>1;
            if(check(mid, n_len, ssize+1)){
                L = mid+1;
                ans = mid;
            }
            else R = mid-1;
        }
        check(ans, n_len, ssize+1);

        if(book) puts("");
        if(ans == 0) puts("?");
        else{
            int len = ans_id.size();
//            printf("%d
", len);
            for(int i = 0; i < len; i++){
                for(int k = ans_id[i]; k-ans_id[i]+1 <= ans; k++){
                    printf("%c", r[k]-100);
                }
                puts("");
            }
        }
        if(!book) book = true;
    }
    return 0;
}