| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 9941 | Accepted: 2740 |
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 ?
題意：給定n 個字符串，求出現在不小於k 個字符串中的最長子串。 思路：將n個字符串連起來，中間用不相同的且沒有出現在字符串中的字符隔開，求後綴數組。然後二分答案，將後綴分成若干組，判斷每組的後綴是否出現在不小於k 個的原串中。這個做法的時間複雜度爲O(nlogn)。
AC代碼：
#include <cstdio>
#include <cstring>
#include <iostream>
#include <cmath>
#include <algorithm>
#include <queue>
#define ll long long
using namespace std;
const int maxn = 200005;
const int INF = 1e9;
int s[maxn];
int sa[maxn], t[maxn], t2[maxn], c[maxn], n, mm;
int rank[maxn], height[maxn];
void build_sa(int n, int m){
int i, *x = t, *y = t2;
for(i = 0; i < m; i++) c[i] = 0;
for(i = 0; i < n; i++) c[x[i] = s[i]]++;
for(i = 1; i < m; i++) c[i] += c[i - 1];
for(i = n - 1; i >= 0; i--) sa[--c[x[i]]] = i;
for(int k = 1; k <= n; k <<= 1)
{
int p = 0;
for(i = n - k; i < n; i++) y[p++] = i;
for(i = 0; i < n; i++) if(sa[i] >= k) y[p++] = sa[i] - k;
for(i = 0; i < m; i++) c[i] = 0;
for(i = 0; i < n; i++) c[x[y[i]]]++;
for(i = 1; i < m; i++) c[i] += c[i - 1];
for(i = n - 1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i];
swap(x, y);
p = 1; x[sa[0]] = 0;
for(i = 1; i < n; i++)
x[sa[i]] = y[sa[i - 1]] == y[sa[i]] && y[sa[i - 1] + k] == y[sa[i] + k] ? p - 1 : p++;
if(p >= n) break;
m = p;
}
}
void getHeight(){
int i, j, k = 0;
for(i = 1; i <= n; i++) rank[sa[i]] = i;
for(i = 0; i < n; i++)
{
if(k) k--;
j = sa[rank[i] - 1];
while(s[i + k] == s[j + k]) k++;
height[rank[i]] = k;
}
}
int rmq[maxn][20];
void initRMQ(){
for(int i = 1; i <= n; i++) rmq[i][0] = rmq[i][0] = height[i];
for(int k = 1; (1 << k) <= n; k++)
for(int i = 1; i + (1 << k) - 1 <= n; i++)
{
rmq[i][k]=min(rmq[i][k-1],rmq[i+(1<<(k-1))][k-1]);
}
}
int lcp(int a,int b)
{
a = rank[a], b = rank[b];
if(a > b) swap(a, b);
a++;
int k = log(b - a + 1.0) / log(2.0);
return min(rmq[a][k], rmq[b - (1 << k) + 1][k]);
}
char str[105][1005];
int id[maxn];
bool ok(int d, bool out){
bool vis[105];
memset(vis, false, sizeof(vis));
int cnt = 0;
for(int i = 1; i <= n; i++)
{
if(height[i] < d)
{
if(cnt > mm / 2 && out)
{
for(int j = 0, k = sa[i - 1]; j < d; j++) printf("%c", s[k + j] + 'a' - 1);
puts("");
}
cnt = 0;
memset(vis, false, sizeof(vis));
}
else
{
if(!vis[id[sa[i - 1]]])
{
vis[id[sa[i - 1]]] = true;
cnt++;
}
if(!vis[id[sa[i]]])
{
vis[id[sa[i]]] = true;
cnt++;
}
if(!out && cnt > mm / 2) return true;
}
}
return false;
}
int main()
{
bool flag = 0;
while(scanf("%d", &mm), mm)
{
if(flag) puts("");
flag = 1;
for(int i = 0; i < mm; i++)
scanf("%s", str[i]);
n = 0;
for(int i = 0 ; i < mm; i++)
{
for(int j = 0; str[i][j]; j++)
{
int ch = str[i][j] - 'a' + 1;
s[n] = ch;
id[n++] = i;
}
s[n] = 27 + i;
id[n++] = i;
}
n--;
s[n] = 0;
build_sa(n + 1, 256);
getHeight();
int low = 0, high = 1000, ans = -1;
while(low <= high)
{
int mid = (low + high) >> 1;
if(ok(mid, 0))
{
ans = mid;
low = mid + 1;
}
else high = mid - 1;
}
if(ans <= 0) printf("?\n");
else ok(ans, 1);
}
return 0;
}