POJ3294:Life Forms——后缀数组

题面

　　POJ3294

解析

　　 翻译一下题意：给定n个字符串，求最长的字符串 S 的长度，使得 S 至少为其中 n/2+1 个字符串的子串, 并输出所有的S，若不存在，则输出'?'。(n<=100)

　　 此题与POJ3261类似(我的博客)，都是相同套路，后缀数组+二分答案

　　当然是把每个字符串连在一起，组成一个大的字符串，再在上面建立后缀数组。但是这里有个细节，搞了我很久，为了防止我们选出的子串跨越了原来的多个字符串，需要在每个字符串中间加入不同的不会出现的字符，这样就解决了这个问题。

　　check的时候需要维护每个块内的$lcp$出现在多少个原来的字符串中，开一个$bool$型的$vis$数组存原字符串是否在当前块内出现，再开一个$pos$数组存长字符串的下标所属的原字符串编号，对于字符串间加入的字符的$pos$等于0， 而$vis[0]$应永远等于$true$，不能对答案产生贡献

　代码：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 105, maxm = 1005;

int n, len, cur, timer, pos[maxn*maxm], s[maxn*maxm];
char c[maxm];

int sa[maxn*maxm], rk[maxn*maxm], tong[maxn*maxm], fir[maxn*maxm], sec[maxn*maxm], hei[maxn*maxm];

void Build_SA()
{
    int m = 150;
    for(int i = 1; i <= len; ++i)
        fir[i] = s[i];
    for(int i = 0; i <= m; ++i)
        tong[i] = 0;
    for(int i = 1; i <= len; ++i)
        tong[fir[i]] ++;
    for(int i = 1; i <= m; ++i)
        tong[i] += tong[i-1];
    for(int i = len; i; --i)
        sa[tong[fir[i]]--] = i;
    for(int k = 1; k <= len; k <<= 1)
    {
        int t = 0;
        for(int i = len - k + 1; i <= len; ++i)
            sec[++t] = i;
        for(int i = 1; i <= len; ++i)
            if(sa[i] - k > 0)
                sec[++t] = sa[i] - k;
        for(int i = 0; i <= m; ++i)
            tong[i] = 0;
        for(int i = 1; i <= len; ++i)
            tong[fir[sec[i]]] ++;
        for(int i = 1; i <= m; ++i)
            tong[i] += tong[i-1];
        for(int i = len; i; --i)
            sa[tong[fir[sec[i]]]--] = sec[i], sec[i] = 0;
        swap(fir, sec);
        t = 0;
        fir[sa[1]] = ++t;
        for(int i = 2; i <= len; ++i)
            if(sec[sa[i]] != sec[sa[i-1]] || sec[sa[i]+k] != sec[sa[i-1]+k])
                fir[sa[i]] = ++t;
            else
                fir[sa[i]] = t;
        if(t >= len)
            break;
        m = t;
    }
}

void Get_hei()
{
    int h = 0;
    for(int i = 1; i <= len; ++i)
        rk[sa[i]] = i;
    for(int i = 1; i <= len ;++i)
    {
        int t = sa[rk[i]-1];
        while(s[t+h] == s[i+h])        h++;
        hei[rk[i]] = h;
        h = max(0, h-1);
    }
}

bool vis[maxn];

bool check(int x)
{
    for(int i = 1; i <= n; ++i)
        vis[i] = 0;
    int p = 1, num = 1;
    vis[0] = 1;
    vis[pos[sa[1]]] = 1;
    for(int i = 2; i <= len; ++i)
    {
        if(hei[i] < x)
        {
            if(num > n / 2)
                return 1;
            for(int j = p; j < i; ++j)
                vis[pos[sa[j]]] = 0;
            vis[0] = 1;
            p = i;
            num = 0;
        }
        if(!vis[pos[sa[i]]])
            num ++, vis[pos[sa[i]]] = 1;
    }
    return num > n / 2;
}

void write(int x)
{
    for(int i = 1; i <= n; ++i)
        vis[i] = 0;
    int p = 1, num = 1;
    vis[0] = 1;
    vis[pos[sa[1]]] = 1;
    for(int i = 2; i <= len; ++i)
    {
        if(hei[i] < x)
        {
            if(num > n / 2)
            {
                int cnt = 0, j = sa[i-1];
                while(cnt < x)
                {
                    if(pos[j] != 0)
                        printf("%c", s[j]-maxn+'a'), cnt++;
                    j++;
                }
                printf("\n");
            }
            for(int j = p; j < i; ++j)
                vis[pos[sa[j]]] = 0;
            vis[0] = 1;
            p = i;
            num = 0;
        }
        if(!vis[pos[sa[i]]])
            num ++, vis[pos[sa[i]]] = 1;
    }
    if(num > n / 2)
    {
        int cnt = 0, j = sa[len];
        while(cnt < x)
        {
            if(pos[j] != 0)
                printf("%c", s[j]-maxn+'a'), cnt++;
            j++;
        }
        printf("\n");
    }
}

void work()
{
    int l = 0, r = len, mid, ans = 0;
    while(l <= r)
    {
        mid = (l + r)>>1;
        if(check(mid))
            ans = mid, l = mid + 1;
        else
            r = mid - 1;
    }
    if(ans)
        write(ans);
    else
        printf("?\n");
}

int main()
{
    while(scanf("%d", &n), n != 0)
    {
        if(cur++)
            printf("\n");
        len = timer = 0;
        for(int i = 1; i <= n; ++i)
        {
            scanf("%s", c);
            int l = strlen(c);
            for(int j = 0; j < l; ++j)
                s[++len] = c[j] - 'a' + maxn, pos[len] = i;
            s[++len] = ++timer;
            pos[len] = 0;
        }
        if(n == 1)
        {
            for(int i = 1; i < len; ++i)
                printf("%c", s[i]-maxn+'a');
            printf("\n");
        }
        else
        {
            Build_SA();
            Get_hei();
            work();
        }
    }
    return 0;
}

 

来源：https://www.cnblogs.com/Joker-Yza/p/11342623.html