POJ 1226 后缀数组 或 KMP 或 暴力

简略题意：出现或反转后出现在每个字符串中的最长字符串。

先将每个串和自己的反转串连接起来，随后将这若干个串连接起来。二分答案，判定就分组看是否有一组后缀在所有原串或者原串的反转串中出现。

#include <iostream>#include <cstring>#include <map>#include <cstdio>#include <vector>#include <algorithm>using namespace std;const int N = 1e5+1100;int n, q;int str[N];char str2[N];int belong[N];int f = 0;int minv;namespace SA {    int sa[N], rank[N], height[N], s[N<<1], t[N<<1], p[N], cnt[N], cur[N];    #define pushS(x) sa[cur[s[x]]--] = x    #define pushL(x) sa[cur[s[x]]++] = x    #define inducedSort(v) fill_n(sa, n, -1); fill_n(cnt, m, 0);                  \        for (int i = 0; i < n; i++) cnt[s[i]]++;                                  \        for (int i = 1; i < m; i++) cnt[i] += cnt[i-1];                           \        for (int i = 0; i < m; i++) cur[i] = cnt[i]-1;                            \        for (int i = n1-1; ~i; i--) pushS(v[i]);                                  \        for (int i = 1; i < m; i++) cur[i] = cnt[i-1];                            \        for (int i = 0; i < n; i++) if (sa[i] > 0 &&  t[sa[i]-1]) pushL(sa[i]-1); \        for (int i = 0; i < m; i++) cur[i] = cnt[i]-1;                            \        for (int i = n-1;  ~i; i--) if (sa[i] > 0 && !t[sa[i]-1]) pushS(sa[i]-1)    void sais(int n, int m, int *s, int *t, int *p) {        int n1 = t[n-1] = 0, ch = rank[0] = -1, *s1 = s+n;        for (int i = n-2; ~i; i--) t[i] = s[i] == s[i+1] ? t[i+1] : s[i] > s[i+1];        for (int i = 1; i < n; i++) rank[i] = t[i-1] && !t[i] ? (p[n1] = i, n1++) : -1;        inducedSort(p);        for (int i = 0, x, y; i < n; i++) if (~(x = rank[sa[i]])) {            if (ch < 1 || p[x+1] - p[x] != p[y+1] - p[y]) ch++;            else for (int j = p[x], k = p[y]; j <= p[x+1]; j++, k++)                if ((s[j]<<1|t[j]) != (s[k]<<1|t[k])) {ch++; break;}            s1[y = x] = ch;        }        if (ch+1 < n1) sais(n1, ch+1, s1, t+n, p+n1);        else for (int i = 0; i < n1; i++) sa[s1[i]] = i;        for (int i = 0; i < n1; i++) s1[i] = p[sa[i]];        inducedSort(s1);    }    template<typename T>    int mapCharToInt(int n, const T *str) {        int m = *max_element(str, str+n);        fill_n(rank, m+1, 0);        for (int i = 0; i < n; i++) rank[str[i]] = 1;        for (int i = 0; i < m; i++) rank[i+1] += rank[i];        for (int i = 0; i < n; i++) s[i] = rank[str[i]] - 1;        return rank[m];    }    template<typename T>    void suffixArray(int n, const T *str) {        int m = mapCharToInt(++n, str);        sais(n, m, s, t, p);        for (int i = 0; i < n; i++) rank[sa[i]] = i;        for (int i = 0, h = height[0] = 0; i < n-1; i++) {            int j = sa[rank[i]-1];            while (i+h < n && j+h < n && s[i+h] == s[j+h]) h++;            if (height[rank[i]] = h) h--;        }    }    template <typename T>    void init(T *str){        str[n] = 0;        suffixArray(n, str);    }    int vis[110], count = 0;    void checkinit() {        memset(vis, 0, sizeof vis);        count = 0;    }    bool check(int x) {        checkinit();        for(int i =  1; i <= n; i++) {            if(height[i] >= x) {                int bl = belong[sa[i]];                if(!vis[bl]) vis[bl] = 1, count ++;                if(count == q)                    return 1;                } else {                    checkinit();                    int bl = belong[sa[i]];                    if(!vis[bl]) vis[bl] = 1, count ++;                }        }        return 0;    }    void solve() {        int l = 0, r = minv + 1, m;        while(l < r) {            m = (l + r) >> 1;            if(check(m)) l = m + 1;            else r = m;        }        printf("%d\n", l - 1);    }};int t;int main() {    scanf("%d", &t);    while(t--) {        scanf("%d", &q);        int s = 0;        int d = 128;        minv = 0x3f3f3f3f;        for(int i = 1; i <= q; i++) {            scanf("%s", str2);            int len = strlen(str2);            for(int j = 0; j < len; j++) str[s+j] = str2[j], belong[s+j] = i;            s += len;            minv = min(minv, len);            str[s] = d;            d++;            s++;            for(int j = len - 1; j >= 0; j--) str[s+len-1-j] = str2[j], belong[s+j] = i;            s += len;            minv = min(minv, len);            str[s] = d;            d++;            s++;        }        n = s;        SA::init(str);        SA::solve();    }    return 0;}