POJ1226 Substrings ——后缀数组 or 暴力+strstr()函数 最长公共子串

题目链接：https://vjudge.net/problem/POJ-1226

 

Substrings

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 15122		Accepted: 5309

Description

You are given a number of case-sensitive strings of alphabetic characters, find the largest string X, such that either X, or its inverse can be found as a substring of any of the given strings.

Input

The first line of the input contains a single integer t (1 <= t <= 10), the number of test cases, followed by the input data for each test case. The first line of each test case contains a single integer n (1 <= n <= 100), the number of given strings, followed by n lines, each representing one string of minimum length 1 and maximum length 100. There is no extra white space before and after a string.

Output

There should be one line per test case containing the length of the largest string found.

Sample Input

2
3
ABCD
BCDFF
BRCD
2
rose
orchid

Sample Output

2
2 

Source

Tehran 2002 Preliminary

 

 

题意：

给出n个字符串，问是否存在一个公共子串存在于每个字符串或其逆串中，若存在，输出最长的长度。

 

题解：

1.将所有字符串已经其逆串拼接在一起，相邻两个之间用各异的分隔符隔开。

2.求出新串的后缀数组，然后二分公共子串的长度mid：mid将新串的后缀分成若干组，每一组对应着一个公共子串，并且长度>=mid。如果这组出现了所有字符串，那么说明当前mid合法，否则不合法。最终求得答案。

3.由于数据较弱，还可以直接暴力+kmp/strstr()函数。

 

后缀数组：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <set>
using namespace std;
typedef long long LL;
const int INF = 2e9;
const LL LNF = 9e18;
const int MOD = 1e9+7;
const int MAXN = 1e5+100;

int id[MAXN];
int r[MAXN], sa[MAXN], Rank[MAXN], height[MAXN];
int t1[MAXN], t2[MAXN], c[MAXN];

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

void DA(int str[], int sa[], int Rank[], int height[], int n, int m)
{
    n++;
    int i, j, p, *x = t1, *y = t2;
    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 <<= 1)
    {
        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(i = 1; i<n; i++)
            x[sa[i]] = cmp(y, sa[i-1], sa[i], j)?p-1:p++;

        if(p>=n) break;
        m = p;
    }

    int k = 0;
    n--;
    for(i = 0; i<=n; i++) Rank[sa[i]] = i;
    for(i = 0; i<n; i++)
    {
        if(k) k--;
        j = sa[Rank[i]-1];
        while(str[i+k]==str[j+k]) k++;
        height[Rank[i]] = k;
    }
}

bool vis[110];
bool test(int n, int len, int k)
{
    int cnt = 0;
    memset(vis, false, sizeof(vis));
    for(int i = 2; i<=len; i++)
    {
        if(height[i]<k)
        {
            cnt = 0;
            memset(vis, false, sizeof(vis));
        }
        else
        {
            if(!vis[id[sa[i-1]]]) vis[id[sa[i-1]]] = true, cnt++;
            if(!vis[id[sa[i]]]) vis[id[sa[i]]] = true, cnt++;
            if(cnt==n) return true;
        }
    }
    return false;
}

char str[MAXN];
int main()
{
    int T, n;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d", &n);
        int len = 0;
        for(int i = 0; i<n; i++)
        {
            scanf("%s", str);
            int LEN = strlen(str);
            for(int j = 0; j<LEN; j++)
            {
                r[len] = str[j];
                id[len++] = i;
            }
            r[len] = 130+i; //分隔符
            id[len++] = i;
        }
        for(int i = 0; i<len; i++)  //逆串
        {
            r[2*len-1-i] = r[i];
            if(r[2*len-1-i]>=130) r[2*len-1-i] += 120;  //分隔符应各异
            id[2*len-1-i] = id[i];
        }
        len *= 2;
        r[len] = 0;
        DA(r,sa,Rank,height,len,400);

        int l = 0, r = 100;
        while(l<=r)
        {
            int mid = (l+r)>>1;
            if(test(n,len,mid))
                l = mid + 1;
            else
                r = mid - 1;
        }
        printf("%d\n", r);
    }
}

 

暴力 + strstr()函数：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <sstream>
#include <algorithm>
using namespace std;
typedef long long LL;
const double eps = 1e-6;
const int INF = 2e9;
const LL LNF = 9e18;
const int MOD = 1e9+7;
const int MAXN = 100+10;

char s[MAXN][MAXN], t1[MAXN], t2[MAXN];

int main()
{
    int T, n;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d", &n);
        for(int i = 1; i<=n; i++)
            scanf("%s", s[i]);

        int Len = strlen(s[1]), ans = 0;
        for(int len = Len; len>=1; len--)
        {
            for(int st = 0; st<=Len-len; st++)
            {
                int en = st+len-1, cnt = 0;
                for(int i = st; i<=en; i++)
                {
                    t1[cnt] = s[1][i];
                    t2[cnt++] = s[1][i];
                }
                t1[cnt] = t2[cnt] = 0;
                reverse(t2,t2+cnt);

                bool flag = true;
                for(int i = 2; i<=n; i++)
                    flag = flag&&(strstr(s[i], t1) || strstr(s[i], t2) );

                if(flag)
                {
                    ans = len;
                    break;
                }
            }
            if(ans==len) break;
        }
        printf("%d\n", ans);
    }
}