POJ 3294 UVA 11107 Life Forms 后缀数组

相同的题目，输出格式有区别。

给定n个字符串，求最长的子串，使得它同时出现在一半以上的串中。

不熟悉后缀数组的童鞋建议先去看一看如何用后缀数组计算两个字符串的最长公共子串 Ural1517

这道题的思路也是基本相同的，都是利用了后缀数组的良好性质。

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;  

const int MAX = 100500;
const int nMAX = 105;
const int mMAX = 1005;  

int strnum;
char str[nMAX][mMAX];
int source[MAX];
int sa[MAX], rk[MAX], height[MAX];
int wa[MAX], wb[MAX], wv[MAX], wd[MAX];
bool vis[nMAX];
int id[MAX];
int anslen, anspos[mMAX], ansnum;
  const int MAXN=200000+100;
void radix(int *str,int *a,int *b,int n,int m)
{
 static int count[MAXN];
 memset(count,0,sizeof(count));
 for(int i=0;i<n;++i)++count[str[a[i]]];
 for(int i=1;i<=m;++i)count[i]+=count[i-1];
 for(int i=n-1;i>=0;--i)b[--count[str[a[i]]]]=a[i];
}

void sorted_suffix_array(int *str,int *sa,int n,int m)
{
 static int rank[MAXN],a[MAXN],b[MAXN];
 for(int i=0;i<n;++i)rank[i]=i;
 radix(str,rank,sa,n,m);

 rank[sa[0]]=0;
 for(int i=1;i<n;++i)rank[sa[i]]=rank[sa[i-1]]+(str[sa[i]]!=str[sa[i-1]]);
 for(int i=0;(1<<i) <n;++i)
 	{
 	 for(int j=0;j<n;++j)
 	 	{
 	 	 a[j]=rank[j]+1;
 	 	 b[j]=j+(1<<i)>=n? 0:rank[j+(1<<i)]+1;
 	 	 sa[j]=j;
		}
	 radix(b,sa,rank,n,n);
	 radix(a,rank,sa,n,n);
	 rank[sa[0]]=0;
	 for(int j=1;j<n;++j)
	 	{
	 	 rank[sa[j]]=rank[sa[j-1]]+(a[sa[j-1]]!=a[sa[j]]||b[sa[j-1]]!=b[sa[j]]);
		}
	}
}

void calc_height(int *str,int *sa,int *h,int n)
{
 static int Rank[MAXN];
 int k=0;
 h[0]=0;
 for(int i=0;i<n;++i)Rank[sa[i]]=i;
 for(int i=0;i<n;++i)
 	{
 	 k= k==0?0:k-1;
 	 if(Rank[i]!=0)
 	   while(str[i+k]==str[sa[Rank[i]-1]+k])++k;
 	 h[Rank[i]]=k;
	}
}

bool solve(int beg, int end)
{
    int tot = 0;
    int t = strnum >> 1;
    for (int i = 0; i < strnum; ++i) vis[i] = false;
    for (int i = beg; i <= end; ++i)
    {
        if (!vis[id[sa[i]]])
        {
            vis[id[sa[i]]] = true;
            ++tot;
        }
        if (tot > t) return true;
    }
    return false;
}  

bool group(int len, int n)
{
    bool res = false;
    int beg, end;
    beg = end = 0;
    for (int i = 1; i < n; ++i)
    {
        if (height[i] >= len) ++end;
        else
        {
            if (solve(beg, end))
            {
                if (!res) ansnum = 0;
                res = true;
                anspos[ansnum++] = sa[beg];
            }
            beg = end = i;
        }
    }
    if (beg < end)
    {
        if (solve(beg, end))
        {
            if (!res) ansnum = 0;
            res = true;
            anspos[ansnum++] = sa[beg];
        }
    }
    return res;
}  

int main()
{ // freopen("t.txt","r",stdin);
 bool flg=false;
    while (scanf("%d", &strnum) && strnum != 0)
    {   if(flg)printf("\n");
        for (int i = 0; i < strnum; ++i) scanf("%s", str[i]);
        int n = 0, low = 1, high = 0, mid;
        for (int i = 0; i < strnum; ++i)
        {
            int j;
            for (j = 0; str[i][j] != 0; ++j)
            {
                id[n] = i;
                source[n++] = str[i][j] - 'a' + 100;
            }
            if (j > high) high = j;
            id[n] = i;
            source[n++] = i;
        }
        sorted_suffix_array(source,sa,n,126);
        calc_height(source,sa,height,n);
        //suffix(source, n, 126);
        //calheight(source, n - 1);
        anslen = 0;
        while (low <= high)
        {
            mid = (low + high) >> 1;
            if (group(mid, n))
            {
                anslen = mid;
                low = mid + 1;
            }
            else high = mid - 1;
        }
        if (anslen == 0) printf("?\n");
        else
        {
            for (int i = 0; i < ansnum; ++i)
            {
                for (int j = 0; j < anslen; ++j)
                {
                    printf("%c", source[anspos[i] + j] - 100 + 'a');
                }
                printf("\n");
            }
        }
        //printf("\n");
		flg=true;
    }
    return 0;
}