poj 3294(二分+后缀数组)

poj 3294

和这题差不多，二分后后缀数组$height$判断，此题要输出所有的解，用个数组存下每个解在$a$中的起始位置即可。不同的是，此题判断时一定要找到一个$height[i]<x$或者循环完毕$height$才能更新解，这样才能防止重复解出现。
(ps: $vi$数组不要开大了，否则memset时容易TLE)

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#define ms(i, j) memset(i, j, sizeof i)
#define FN2 "poj3296" 
using namespace std;

const int MAXN = 105 * 1005 + 5;

char s[MAXN];
int ans,ba,bf[MAXN],mini,t,n,m,a[MAXN],tp[MAXN],SA[MAXN],rk[MAXN],tax[MAXN],height[MAXN],belong[MAXN];

bool cmp(int *f, int i, int k) {return f[SA[i]]==f[SA[i-1]]&&f[SA[i]+k]==f[SA[i-1]+k];}
void build() {
    for (int i=0;i<m;i++) tax[i] = 0;
    for (int i=0;i<n;i++) tax[rk[i]=a[i]]++;
    for (int i=1;i<m;i++) tax[i] += tax[i-1];
    for (int i=n-1;i>=0;i--) SA[--tax[rk[i]]] = i;
    int p;
    for (int k=1;k<=n;k*=2) {
        p = 0;
        for (int i=n-k;i<n;i++) tp[p++] = i;
        for (int i=0;i<n;i++) if (SA[i]>=k) tp[p++] = SA[i] - k;

        for (int i=0;i<m;i++) tax[i] = 0;
        for (int i=0;i<n;i++) tax[rk[tp[i]]]++;
        for (int i=1;i<m;i++) tax[i] += tax[i-1];
        for (int i=n-1;i>=0;i--) SA[--tax[rk[tp[i]]]] = tp[i];

        swap(rk, tp), p = 0, rk[SA[0]] = 0;
        for (int i=1;i<n;i++) rk[SA[i]] = cmp(tp, i, k) ? p : ++p;
        if (++p>=n) break;
        m = p;
    }
}
void getH() {
    int k = 0;
    for (int i=0;i<n;i++) {
        if (k) k--;
        int j = SA[rk[i]-1];
        while (a[i+k]==a[j+k]) k++;
        height[rk[i]] = k;
    }
}
void init() {
    n = 0, m = 200, mini = 200000000;
    for (int i=0;i<t;i++) {
        scanf("%s", s);
        int l = strlen(s);
        mini = min(mini, l);
        for (int j=0;j<l;j++) belong[n] = i, a[n++] = s[j];
        belong[n] = 104; a[n++] = m++;
    }
    a[n-1] = 0, m+=5;
}
int vi[100 + 5];
bool check(int x) {
    int tot = 0, flag = false;
    ms(vi, false);
    for (int i=2;i<n;i++) {
        if (height[i]<x) {
            if (tot>t/2) {
                flag = true;
                if (x>ba) {
                    ans = 0;
                    ba = x;
                    bf[ans] = SA[i-1];
                    ans++;
                } else if (x==ba) {
                    bf[ans] = SA[i-1];
                    ans++;
                }
            }
            ms(vi, false);
            tot = 0;
            continue;
        }
        if (!vi[belong[SA[i]]]) vi[belong[SA[i]]] = true, ++tot;
        if (!vi[belong[SA[i-1]]]) vi[belong[SA[i-1]]] = true, ++tot;
    }
    if (tot>t/2) {
        if (x>ba) {
            ans = 0;
            ba = x;
            bf[ans] = SA[n-1];
            ans++;
        } else if (x==ba) {
            bf[ans] = SA[n-1];
            ans++;
        }
        return true;
    }    
    return flag;
}
void solve() {
    build(), getH();
    int l = 1, r = mini + 1;
    ba = 0, ms(bf, 0), ans = 0;
    while (l<r) {
        int mid = (l+r)/2;
        if (check(mid)) {
            l = mid + 1;
        } else r = mid;
    }
    if (!ans) printf("?\n"); else {
        for (int i=0;i<ans;i++) {
            for (int j=bf[i];j<bf[i]+ba;j++) {
                printf("%c", a[j]);
            }
            printf("\n");
        }
    }
    printf("\n");
}
int main() {
    #ifndef ONLINE_JUDGE
    freopen(FN2".in","r",stdin);freopen("1.out","w",stdout);
    #endif
    while(scanf("%d", &t)==1&&t) init(), solve();
    return 0;
}