UVA11107 Life Forms --- 后缀数组

UVA11107 Life Forms

题目描述：

求出出现在一半以上的字符串内的最长字符串。

数据范围：

\(\sum len(string) <= 10^{5}\)

非常坑的题目。

思路非常好想。

构造出后缀数组。

二分出\(len\)后用\(height\)分组

记\(bel(i)\)表示排名为\(i\)后缀属于哪一个串

当同一组内的不同的\(bel(i)\)出现了\(n/2\)时，本组内有一组解。

注意：

每行数据间要打一个空行

#include <cstdio>
#include <cstring>
#include <iostream>
#define sid 200050
#define ri register int
using namespace std;
 
template <typename re>
inline void upmax(re &a, re b) { if(a < b) a = b; }
 
int Tt, n, ml, ned;
char s[sid];
int bel[sid], sc[sid];
int flag[];
int sa[sid], rk[sid], cnt[sid], p1[sid], p2[sid], ht[sid];
 
inline void Suffix() {
    int m = ;
    int *t1 = p1, *t2 = p2;
    for(ri i = ; i <= n; i ++) sc[i] = (s[i] == ) ? ++ m : s[i];
    for(ri i = ; i <= n; i ++) t1[i] = sc[i];
    for(ri i = ; i <= m; i ++) cnt[i] = ;
    for(ri i = ; i <= n; i ++) cnt[t1[i]] ++;
    for(ri i = ; i <= m; i ++) cnt[i] += cnt[i - ];
    for(ri i = n; i >= ; i --) sa[cnt[t1[i]] --] = i;
    for(ri k = ; k <= n; k <<= ) {
        int p = ;
        for(ri i = ; i <= m; i ++) t2[i] = ;
        for(ri i = n - k + ; i <= n; i ++) t2[++ p] = i;
        for(ri i = ; i <= n; i ++) if(sa[i] > k) t2[++ p] = sa[i] - k;
        for(ri i = ; i <= m; i ++) cnt[i] = ;
        for(ri i = ; i <= n; i ++) cnt[t1[t2[i]]] ++;
        for(ri i = ; i <= m; i ++) cnt[i] += cnt[i - ];
        for(ri i = n; i >= ; i --) sa[cnt[t1[t2[i]]] --] = t2[i];
        swap(t1, t2); t1[sa[]] = p = ;
        for(ri i = ; i <= n; i ++)
        t1[sa[i]] = (t2[sa[i]] == t2[sa[i - ]] && t2[sa[i] + k] == t2[sa[i - ] + k]) ? p : ++ p;
        m = p; if(p >= n) break;
    }
    for(ri i = ; i <= n; i ++) rk[sa[i]] = i;
    ri k = , j;
    for(ri i = ; i <= n; i ++) {
        if(k) k --;
        j = sa[rk[i] - ];
        while(sc[j + k] == sc[i + k]) k ++;
        ht[rk[i]] = k;
    }
}
 
inline bool Check(int htk) {
    int cnt = , tim = ;
    memset(flag, , sizeof(flag));
    for(ri i = ; i <= n; i ++) {
        if(ht[i] < htk) cnt = , ++ tim;
        if(flag[bel[sa[i]]] != tim && bel[sa[i]]) cnt ++, flag[bel[sa[i]]] = tim;
        if(cnt >= ned) return ;
    }
    return ;
}
 
inline int Binary() {
    int l = , r = ml, ans = -;
    while(l <= r) {
        int mid = (l + r) >> ;
        if(Check(mid)) l = mid + , ans = mid;
        else r = mid - ;
    }
    return ans;
}
 
inline void Get_Ans(int op) {
    if(op == -) {
        printf("?\n");
        return;
    }
    memset(flag, , sizeof(flag));
    int cnt = , tim = , fag;
    for(ri i = ; i <= n; i ++) {
        if(ht[i] < op) fag = , cnt = , ++ tim;
        if(flag[bel[sa[i]]] != tim && bel[sa[i]]) cnt ++, flag[bel[sa[i]]] = tim;
        if(cnt >= ned && !fag) {
            int k = sa[i];
            for(ri j = ; j <= op; j ++) printf("%c", s[k + j - ]);
            printf("\n");
            cnt = ; fag = ;
        }
    }
}
 
int main() {
    bool pe = ;
    while(scanf("%d", &Tt) ==  && Tt) {
        if(pe) printf("\n"); pe = ;
        n = ml = ; ned = Tt /  + ;
        memset(s, , sizeof(s));
        memset(bel, , sizeof(bel));
        for(ri i = ; i <= Tt; i ++) {
            scanf("%s", s +  + n);
            int nl = strlen(s +  + n);
            for(ri j = n + ; j <= n + nl; j ++) bel[j] = i;
            upmax(ml, nl); n += nl; s[++ n] = ;
        }
        if(Tt != ) {
            Suffix();
            int ans = Binary();
            Get_Ans(ans);
        }
        else {
            for(ri j = ; j <= n - ; j ++) printf("%c", s[j]);
            printf("\n");
        }
    }
    return ;
}

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