Life Forms POJ - 3294（不小于k个字符串中的最长子串）

题意：

　　求不小于字符串一半长度个字符串中的最长字串

解析：

　　论文题例11

　　将n个字符串连起来，中间用不相同的且没有出现在字符串中的字符隔开， 求后缀数组， 然后二分答案变为判定性问题，

然后判断每组的后缀是否出现在不小于 k 个的原串中， 这个做法的时间复杂度为O（nlogn）

#include <iostream>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <map>
#include <cctype>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#define rap(i, a, n) for(int i=a; i<=n; i++)
#define rep(i, a, n) for(int i=a; i<n; i++)
#define lap(i, a, n) for(int i=n; i>=a; i--)
#define lep(i, a, n) for(int i=n; i>a; i--)
#define rd(a) scanf("%d", &a)
#define rlld(a) scanf("%lld", &a)
#define rc(a) scanf("%c", &a)
#define rs(a) scanf("%s", a)
#define MOD 2018
#define LL long long
#define ULL unsigned long long
#define Pair pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define _  ios_base::sync_with_stdio(0),cin.tie(0)
//freopen("1.txt", "r", stdin);
using namespace std;
const int maxn = 1000005, INF = 0x7fffffff;
int s[maxn];
int sa[maxn], t[maxn], t2[maxn], c[maxn], n;
int ran[maxn], height[maxn], length[105], ans[maxn];
bool vis[105];
void get_sa(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 = 0; 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;
    }
    int k = 0;
    for(i = 0; i < n; i++) ran[sa[i]] = i;
    for(i = 0; i < n; i++)
    {
        if(k) k--;
        int j = sa[ran[i]-1];
        while(s[i+k] == s[j+k]) k++;
        height[ran[i]] = k;
    }
}

int solve(int k, int q)
{
    mem(vis, 0);
    int cnt = 0, siz = 0;
    for(int i=1; i<n; i++)
    {
        if(height[i] < k)
        {
            if(cnt > q/2) ans[++siz] = sa[i-1];
            mem(vis, 0);
            cnt = 0;
            continue;
        }
        for(int j=1; j<=q; j++)
        {
            if(sa[i] < length[j] && sa[i] > length[j-1]) if(!vis[j]) cnt++, vis[j] = 1;
            if(sa[i-1] < length[j] && sa[i-1] > length[j-1]) if(!vis[j]) cnt++, vis[j] = 1;
        }
    }
    if(cnt > q/2) ans[++siz] = sa[n-1];
    if(siz)
    {
        ans[0] = siz;
        return true;
    }
    return false;
}

int q;
char str[maxn];
int main()
{
    int flag = false;
    while(~rd(q) && q)
    {
        length[0]  = 0;
        int l = 1, r = 0, len;
        n = 0;
        int cnt = 28;
        rep(i, 0, q)
        {
            rs(str);
            if(!i) len = strlen(str), r = len;
            rep(j, 0, len)
                s[n++] = str[j] - 'a' + 1;
            length[i+1] = n;
            s[n++] = cnt++;
        }
        s[n++] = 0;
        get_sa(200);

        while(l <= r)
        {
            int mid = l + (r - l) / 2;
            if(solve(mid, q)) l = mid + 1;
            else r = mid - 1;
        }
        if(flag) printf("\n");
        else flag = true;

        if(l == 1)
        {
            printf("?\n");
            continue;
        }
        for(int i=1; i<=ans[0]; i++)
        {
            for(int j=ans[i]; j<=ans[i]+r-1; j++)
                printf("%c", s[j]-1 + 'a');
            printf("\n");
        }
    }
    return 0;
}