POJ 3294Life Forms

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

3abcdefgbcdefghcdefghi3xxxyyyzzz0

Sample Output

bcdefgcdefgh
?
求出现在一半以上的串里的最长子串，wa了好多发，坑点是不同串之间加字母会炸，改成数字好了。
#include<set>#include<map>#include<ctime>#include<cmath>#include<stack>#include<queue>#include<bitset>#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<algorithm>#include<functional>#define rep(i,j,k) for (int i = j; i <= k; i++)#define per(i,j,k) for (int i = j; i >= k; i--)#define lson x << 1, l, mid#define rson x << 1 | 1, mid + 1, r#define fi first#define se second#define mp(i,j) make_pair(i,j)#define pii pair<int,int>using namespace std;typedef long long LL;const int low(int x) { return x&-x; }const double eps = 1e-8;const int INF = 0x7FFFFFFF;const int mod = 1e9 + 7;const int N = 2e5 + 10;const int read(){char ch = getchar();while (ch<'0' || ch>'9') ch = getchar();int x = ch - '0';while ((ch = getchar()) >= '0'&&ch <= '9') x = x * 10 + ch - '0';return x;}int m, T = 0;struct Sa{char ss[N];int s[N];int rk[2][N], sa[N], h[N], w[N], now, n;int rmq[N][20], lg[N], bel[N];bool GetS(){scanf("%d", &m);if (!m) return false;if (!T) T++; else printf("\n");int q = 'z' + 1;n = 0;rep(i, 0, m - 1){scanf("%s", ss + n + 1);for (++n; ss[n]; ++n) bel[n] = i, s[n] = ss[n];s[n] = q++; bel[n] = 100;}s[n--] = 0;return true;}void getsa(int z, int &m){int x = now, y = now ^= 1;rep(i, 1, z) rk[y][i] = n - i + 1;for (int i = 1, j = z; i <= n; i++)if (sa[i] > z) rk[y][++j] = sa[i] - z;rep(i, 1, m) w[i] = 0;rep(i, 1, n) w[rk[x][rk[y][i]]]++;rep(i, 1, m) w[i] += w[i - 1];per(i, n, 1) sa[w[rk[x][rk[y][i]]]--] = rk[y][i];for (int i = m = 1; i <= n; i++){int *a = rk[x] + sa[i], *b = rk[x] + sa[i - 1];rk[y][sa[i]] = *a == *b&&*(a + z) == *(b + z) ? m - 1 : m++;}}void getsa(int m){rk[1][0] = now = sa[0] = s[0] = 0;rep(i, 1, m) w[i] = 0;rep(i, 1, n) w[s[i]]++;rep(i, 1, m) rk[1][i] = rk[1][i - 1] + (bool)w[i];rep(i, 1, m) w[i] += w[i - 1];rep(i, 1, n) rk[0][i] = rk[1][s[i]];rep(i, 1, n) sa[w[s[i]]--] = i;rk[1][n + 1] = rk[0][n + 1] = 0;//多组的时候容易出bugfor (int x = 1, y = rk[1][m]; x <= n && y <= n; x <<= 1) getsa(x, y);for (int i = 1, j = 0; i <= n; h[rk[now][i++]] = j ? j-- : j){if (rk[now][i] == 1) continue;int k = n - max(sa[rk[now][i] - 1], i);while (j <= k && s[sa[rk[now][i] - 1] + j] == s[i + j]) ++j;}}void getrmq(){h[n + 1] = h[1] = lg[1] = 0;rep(i, 2, n) rmq[i][0] = h[i], lg[i] = lg[i >> 1] + 1;for (int i = 1; (1 << i) <= n; i++){rep(j, 2, n){if (j + (1 << i) > n + 1) break;rmq[j][i] = min(rmq[j][i - 1], rmq[j + (1 << i - 1)][i - 1]);}}}int lcp(int x, int y){int l = min(rk[now][x], rk[now][y]) + 1, r = max(rk[now][x], rk[now][y]);return min(rmq[l][lg[r - l + 1]], rmq[r - (1 << lg[r - l + 1]) + 1][lg[r - l + 1]]);}bool check(int x){bitset<101> b;b.reset();rep(i, 2, n){if (h[i] >= x) b[bel[sa[i]]] = b[bel[sa[i - 1]]] = 1;else{b[100] = 0;if (b.count() * 2 > m) return true;b.reset();}}b[100] = 0;return b.count() * 2 > m;}void putout(int x){bitset<101> b;b.reset();rep(i, 2, n){if (h[i] >= x) b[bel[sa[i]]] = b[bel[sa[i - 1]]] = 1;else{b[100] = 0;if (b.count() * 2 > m){rep(j, 1, x) printf("%c", ss[sa[i - 1] + j - 1]);putchar(10);}b.reset();}}b[100] = 0;if (b.count() * 2 > m){rep(j, 1, x) printf("%c", ss[sa[n] + j - 1]);putchar(10);}}void work(){if (m == 1) { printf("%s\n", ss + 1); return; }getsa(300);int l = 1, r = n;while (l <= r){if (check(l + r >> 1)) l = (l + r >> 1) + 1;else r = (l + r >> 1) - 1;}if (l - 1) putout(l - 1); else printf("?\n");}}sa;int main(){while (sa.GetS()) sa.work();return 0;}