Poj 3294 Life Forms (后缀数组 + 二分 + Hash)

题目链接:

　　Poj 3294 Life Forms

题目描述：

　　有n个文本串，问在一半以上的文本串出现过的最长连续子串？

解题思路：

　　可以把文本串用没有出现过的不同字符连起来，然后求新文本串的height。然后二分答案串的长度K，根据K把新文本串的后缀串分块，统计每块中的原文本串出现的次数，大于原文本串数目的一半就作为答案记录下来，对于输出字典序，height就是排好序的后缀数组，只要按照顺序输出即可。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
const int maxn = 110000;

int sa[maxn], rank[maxn], height[maxn], vis[110], res[maxn];
int t1[maxn], t2[maxn], r[maxn], flag[maxn], c[maxn];

bool cmp (int *str, int a, int b, int k)
{
    return str[a]==str[b] && str[a+k]==str[b+k];
}

void da (int *str, int n, int m)
{
    n ++;
    int *x = t1, *y = t2, i, j;
    
    for (i=0; i<m; i++) c[i] = 0;
    for (i=0; i<n; i++) c[x[i]=str[i]] ++;
    for (i=1; i<m; i++) c[i] += c[i-1];
    for (i=n-1; i>=0; i--) sa[-- c[x[i]]] = i;
    
    for (j=1; j<=n; j*=2)
    {
        int p = 0;
        for (i=n-j; i<n; i++) y[p++] = i;
        for (i=0; i<n; i++) if (sa[i] >= j) y[p++] = sa[i] - j;

        for (i=0; i<m; i++) c[i] = 0;
        for (i=0; i<n; i++) c[x[y[i]]] ++;
        for (i=1; i<m; i++) c[i] += c[i-1];
        for (i=n-1; i>=0; i--) sa[-- c[x[y[i]]]] = y[i];

        swap (x, y);
        p = 1;
        x[sa[0]] = 0;
        for (int i=1; i<n; i++)//i是rank
            x[sa[i]] = cmp(y, sa[i-1], sa[i], j)?p-1:p++;
        if (p >= n)
            break;
        m = p;
    }
    
    for (i=1; i<n; i++)
        rank[sa[i]] = i;
        
    int k = 0;
    n --;
    for (int i=0; i<n; i++)
    {
        if (k)  k --;
        int j = sa[rank[i] - 1];
        while (str[i+k] == str[j+k])    k++;
        height[rank[i]] = k;
    }
}

bool Bin_sreach (int x, int k, int n)
{
    int ans, num;
    ans = num = 0;
    memset (vis, 0, sizeof(vis));
    
    for (int i=2; i<=k; i++)
    {
        if (height[i] >= x)
        {
            ans += vis[flag[sa[i-1]]]?0:1;
            vis[flag[sa[i-1]]] = 1;
            
            ans += vis[flag[sa[i]]]?0:1;
            vis[flag[sa[i]]] = 1;
        }
        else
        {
            if (ans*2 > n)
                res[++ num] = sa[i-1];
                
            ans = 0;
            memset (vis, 0, sizeof(vis));
        }
    }
    if (ans*2 > n)
        res[++ num] = sa[k-1];
        
    if (num)
    {
        res[0] = num;
        return true;
    }
    return false;
}

int main ()
{
    int n, l = 0;
    char str[1010];
    while (scanf ("%d", &n), n)
    {
        if (l ++)
            printf ("
");
            
        int k = 0;
        for (int i=0; i<n; i++)
        {
            scanf ("%s", str);
            for (int j=0; str[j]; j++)
            {
                r[k] = str[j];
                flag[k++] = i;//记录k字母所在的字符串
            }
            r[k] = 130 + i;
            flag[k++] = -1;
        }
        
        r[k] = 0;
        da (r, k, 250);
        
        int low = 1, high = k, mid, ans = 0;
        while (low <= high)
        {//二分枚举
            mid = (low + high) / 2;
            if (Bin_sreach(mid, k, n))
            {
                ans = mid;
                low = mid + 1;
            }
            else
                high = mid - 1;
        }
        
        if (low == 1)
        {
            printf ("?
");
            continue;
        }
        
        for (int i=1; i<=res[0]; i++)
            {
                for (int j=res[i]; j<res[i]+ans; j ++)
                    printf ("%c", r[j]);
                printf ("
");
            }
    }
    return 0;
}