Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

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

3
abcdefg
bcdefgh
cdefghi
3
xxx
yyy
zzz
0

Output for Sample Input

bcdefg
cdefgh

?

---

錯誤的代碼一直過，真令人畏懼。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
struct SuffixArray {
    int sa[110005], h[110005], n;
    char str[110005];

    int w[110005], ta[110005], tb[110005]; // buffer
    void sort(int *x, int *y, int m) { // radix sort
        static int i;
        for(i = 0; i < m; i++)
            w[i] = 0;
        for(i = 0; i < n; i++)
            w[x[y[i]]]++;
        for(i = 1; i < m; i++)
            w[i] += w[i-1];
        for(i = n-1; i >= 0; i--)
            sa[--w[x[y[i]]]] = y[i];
    }
    bool cmp(int *r, int a, int b, int l) {
        if(r[a] == r[b]) {
            if(a+l >= n || b+l >= n)
                return false;
            return r[a+l] == r[b+l];
        }
        return false;
    }
    void build_h() {
        int i, j, k;
        for(i = 0; i < n; i++)  ta[sa[i]] = i;
        for(i = 0; i < n; i++) {
            if(ta[i] == 0) {
                h[ta[i]] = 0;
                continue;
            }
            if(i == 0 || h[ta[i-1]] <= 1)
                k = 0;
            else
                k = h[ta[i-1]]-1;
            while(str[sa[ta[i]-1]+k] == str[sa[ta[i]]+k])
                k++;
            h[ta[i]] = k;
        }
    }
    void build() {// x: rank, y: second key(array index)
        int i, k, m = 128, p;
        int *x = ta, *y = tb, *z;
        n = strlen(str);
        x[n] = 0;
        for(i = 0; i < n; i++)
            x[i] = str[i], y[i] = i;
        sort(x, y, m);
        for(k = 1, p = 1; p < n; k *= 2, 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;
                }
            }
            sort(x, y, m);
            z = x, x = y, y = z;
            for(i = 1, p = 1, x[sa[0]] = 0; i < n; i++)
                x[sa[i]] = cmp(y, sa[i-1], sa[i], k) ? p-1 : p++;
        }
    }
};
SuffixArray SA;
int Sfrom[110005], Wlen[105], n, half;
char mark[105];
int check(int len, int printflag) {
    int i, j, k, cnt;
    for(i = n; i < SA.n; i++) {
        if(SA.h[i] >= len) {
            memset(mark, 0, sizeof(mark));
            j = i;
            cnt = 1;
            mark[Sfrom[SA.sa[i-1]]] = 1;
            while(j < SA.n && SA.h[j] >= len) {
                if(Sfrom[SA.sa[j]] != Sfrom[SA.sa[j]+len-1]) {
                    j++;break;
                }
                if(mark[Sfrom[SA.sa[j]]] == 0) {
                    mark[Sfrom[SA.sa[j]]] = 1;
                    cnt++;
                }
                j++;
            }
            if(cnt > half) {
                if(printflag) {
                    for(k = 0; k < len; k++)
                        putchar(SA.str[SA.sa[i]+k]);
                    puts("");
                } else {
                    return 1;
                }
            }
            i = j-1;
        }
    }
    return 0;
}
int main() {
    int i, first = 0;
    while(scanf("%d", &n) == 1 && n) {
        if(first)   puts("");
        first = 1;
        int m = 0;
        half = n/2; // > half
        int mxlen = 0;
        for(i = 0; i < n; i++) {
            scanf("%s", SA.str+m);
            int cnt = 0;
            while(SA.str[m])    Sfrom[m] = i, m++, cnt++;
            if(cnt > mxlen) mxlen = cnt;
            Wlen[i] = cnt;
            SA.str[m++] = '$';
            SA.str[m] = 0;
        }
        SA.str[m-1] = '\0';
        if(n == 1) {
            puts(SA.str);
            continue;
        }
        //puts(SA.str);
        SA.build();
        SA.build_h();
        int l = 1, r = mxlen;
        int res = 0;
        while(l <= r) {
            m = (l+r)/2;
            if(check(m, 0)) {
                l = m+1;
                if(m > res)
                    res = m;
            } else {
                r = m-1;
            }
        }
        if(res == 0)
            puts("?");
        else
            check(res, 1);
    }
    return 0;
}