poj 3294 Life Forms - 后缀数组 - 二分答案

题目传送门

　　传送门I

　　传送门II

题目大意

　　给定$n$个串，询问所有出现在严格大于$\frac{n}{2}$个串的最长串。不存在输出'?'

　　用奇怪的字符把它们连接起来。然后求sa，hei，二分答案，按mid分组。

　　判断每一组存在的后缀属于的原串的种类数是不是存在那么多个。

　　这个做法可以推广到多串求LCS，然后多个log，完美在SPOJ上T掉。

Code

/**
 * poj
 * Problem#3294
 * Accepted
 * Time: 391ms
 * Memory: 5024k
 */
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#ifndef WIN32
#define Auto "%lld"
#else
#define Auto "%I64d"
#endif
using namespace std;
typedef bool boolean;
#define ll long long

#define pii pair<int, int>
#define fi first
#define sc second

const int N = 1e5 + 105, M = 105;

typedef class Pair3 {
    public:
        int x, y, id;

        Pair3() {    }
        Pair3(int x, int y, int id):x(x), y(y), id(id) {    }
}Pair3;

typedef class SuffixArray {
    protected: 
        Pair3 T1[N], T2[N];
        int cnt[N];

    public:
        int n;
        char *str;
        int sa[N], rk[N], hei[N];

        void set(int n, char* str) {
            this->n = n;
            this->str = str;
            memset(sa, 0, sizeof(sa));
            memset(rk, 0, sizeof(rk));
            memset(hei, 0, sizeof(hei));
        }

        void radix_sort(Pair3* x, Pair3* y) {
            int m = max(n, 256);
            memset(cnt, 0, sizeof(int) * m);
            for (int i = 0; i < n; i++)
                cnt[x[i].y]++;
            for (int i = 1; i < m; i++)
                cnt[i] += cnt[i - 1];
            for (int i = 0; i < n; i++)
                y[--cnt[x[i].y]] = x[i];

            memset(cnt, 0, sizeof(int) * m);
            for (int i = 0; i < n; i++)
                cnt[y[i].x]++;
            for (int i = 1; i < m; i++)
                cnt[i] += cnt[i - 1];
            for (int i = n - 1; ~i; i--)
                x[--cnt[y[i].x]] = y[i];
        }

        void build() {
            for (int i = 0; i < n; i++)
                rk[i] = str[i];
            for (int k = 1; k < n; k <<= 1) {
                for (int i = 0; i + k < n; i++)
                    T1[i] = Pair3(rk[i], rk[i + k], i);
                for (int i = n - k; i < n; i++)
                    T1[i] = Pair3(rk[i], 0, i);
                radix_sort(T1, T2);
                int diff = 0;
                rk[T1[0].id] = 0;
                for (int i = 1; i < n; i++)
                    rk[T1[i].id] = (T1[i].x == T1[i - 1].x && T1[i].y == T1[i - 1].y) ? (diff) : (++diff);
                if (diff == n - 1)
                    break;
            }
            for (int i = 0; i < n; i++)
                sa[rk[i]] = i;
        }

        void get_height() {
            for (int i = 0, j, k = 0; i < n; i++, (k) ? (k--) : (0)) {
                if (rk[i]) {
                    j = sa[rk[i] - 1];
                    while (i + k < n && j + k < n && str[i + k] == str[j + k])    k++;
                    hei[rk[i]] = k;
                }
            }        
        }

        const int& operator [] (int p) {
            return sa[p];
        }

        const int& operator () (int p) {
            return hei[p];
        }
}SuffixArray;

int n, m, K;
char S[N];
int suf[N];
SuffixArray sa;

inline boolean init() {
    scanf("%d", &m);
    if (!m)
        return false;
    K = (m >> 1) + 1;
    scanf("%s", S);
    n = strlen(S);
    for (int i = 1; i < m; i++) {
        S[n++] = '?';
        scanf("%s", S + n);
        n += strlen(S + n);
    }
    suf[n] = 0;
    for (int i = n - 1; ~i; i--)
        suf[i] = suf[i + 1] + (S[i] == '?');
    sa.set(n, S);
    return true;
}

int chk_clock = 0;
int vis[M];

boolean check(int L, int R, int mid) {    // [L, R)
    if (R - L < K)
        return false;
    if (suf[sa[L]] - suf[sa[L] + mid])
        return false;
    chk_clock++;
    int rt = 0;
    for (int i = L; i < R && rt < K; i++)
        if (vis[suf[sa[i]]] != chk_clock)
            vis[suf[sa[i]]] = chk_clock, rt++;
    return rt >= K;
}

boolean check(int mid) {
    int lst = 0;
    boolean rt = false;
    for (int i = 1; i < n && !rt; i++)
        if (sa(i) < mid)
            rt |= check(lst, i, mid), lst = i;
    if (!rt)
        rt |= check(lst, n, mid);
    return rt;
}

void output(int L, int R, int res) {
    if (!check(L, R, res))
        return;
    for (int i = sa[L], j = 0; j < res; j++)
        putchar(S[i + j]);
    putchar('\n');
}

inline void solve() {
    chk_clock = 0;
    memset(vis, 0, sizeof(vis));
    
    sa.build();
    sa.get_height();

    int l = 1, r = n / m, mid;
    while (l <= r) {
        mid = (l + r) >> 1;
        if (check(mid))
            l = mid + 1;
        else
            r = mid - 1;
    }
    
    if (!(--l))
        puts("?");
    else {
        int lst = 0;
        for (int i = 1; i < n; i++)
            if (sa(i) < l)
                output(lst, i, l), lst = i;
        output(lst, n, l);
    }
    puts("");
}

int main() {
    while (init())
        solve();
    return 0;
}