求N個字符串最長的公共子串。這題數據比較水,暴力第一個字符串的子串也可以過。
初學后綴數組,有很多不明白的東西,此題后綴數組的代碼在網上也是一把抓。
說實話我確實還不懂后綴數組,但是后綴數組太強大了,只能硬著頭皮照著葫蘆畫瓢了。
貼下代碼方便以后查閱吧。。。
感覺后綴數組的應用最主要的還是height數組,看懂倍增算法排序后綴已經非常困難了。
然后再理解height數組怎么用也不是一件容易的事情。然后貌似height數組最關鍵的用法是
枚舉某一個長度的子串時候,比如長度為k,能夠用這個k對height數組進行分組,這個羅穗騫
的論文里面有個求不重疊最長重復子串的例子說明了這個height數組分組的思路,不過我現在
還是不怎么理解。。。
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int MAX_N = 110;
const int MAX_L = MAX_N * MAX_N;
char szStr[MAX_N];
int nNum[MAX_L];
int nLoc[MAX_L];
bool bVisit[MAX_N];
int sa[MAX_L], rank[MAX_L], height[MAX_L];
int wa[MAX_L], wb[MAX_L], wv[MAX_L], wd[MAX_L];
int cmp(int* r, int a, int b, int l)
{
return r[a] == r[b] && r[a + l] == r[b + l];
}
//倍增算法,r為待匹配數組,n為總長度,m為字符串范圍
void da(int* r, int n, int m)
{
int i, j, p, *x = wa, *y = wb;
for (i = 0; i < m; ++i) wd[i] = 0;
for (i = 0; i < n; ++i) wd[x[i] = r[i]]++;
for (i = 1; i < m; ++i) wd[i] += wd[i - 1];
for (i = n - 1; i >= 0; --i) sa[--wd[x[i]]] = i;
for (j = 1, p = 1; p < n; j *= 2, m = p)
{
for (p = 0, 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 < n; ++i) wv[i] = x[y[i]];
for (i = 0; i < m; ++i) wd[i] = 0;
for (i = 0; i < n; ++i) wd[wv[i]]++;
for (i = 1; i < m; ++i) wd[i] += wd[i - 1];
for (i = n - 1; i >= 0; --i) sa[--wd[wv[i]]] = y[i];
swap(x, y);
for (p = 1, x[sa[0]] = 0, i = 1; i < n; ++i)
{
x[sa[i]] = cmp(y, sa[i - 1], sa[i], j)? p - 1 : p++;
}
}
}
//求height數組
void calHeight(int* r, int n)
{
int i, j, k = 0;
for (i = 1; i <= n; ++i) rank[sa[i]] = i;
for (i = 0; i < n; height[rank[i++]] = k)
{
if (k) --k;
for(j = sa[rank[i] - 1]; r[i + k] == r[j + k]; k++);
}
}
bool Check(int nMid, int nLen, int nN)
{
int nCnt = 0;
memset(bVisit, false, sizeof(bVisit));
for (int i = 2; i <= nLen; ++i)
{
if (nMid > height[i])
{
nCnt = 0;
memset(bVisit, false, sizeof(bVisit));
continue;
}
if (!bVisit[nLoc[sa[i - 1]]])
{
bVisit[nLoc[sa[i - 1]]] = true;
++nCnt;
}
if (!bVisit[nLoc[sa[i]]])
{
bVisit[nLoc[sa[i]]] = true;
++nCnt;
}
if (nCnt == nN) return true;
}
return false;
}
int main()
{
int nT;
scanf("%d", &nT);
while (nT--)
{
int nN;
int nEnd = 300;
int nP = 0;
scanf("%d", &nN);
for (int i = 1; i <= nN; ++i)
{
scanf("%s", szStr);
char* pszStr;
for (pszStr = szStr; *pszStr; ++pszStr)
{
nLoc[nP] = i;
nNum[nP++] = *pszStr;
}
nLoc[nP] = nEnd;
nNum[nP++] = nEnd++;
reverse(szStr, szStr + strlen(szStr));
for (pszStr = szStr; *pszStr; ++pszStr)
{
nLoc[nP] = i;
nNum[nP++] = *pszStr;
}
nLoc[nP] = nEnd;
nNum[nP++] = nEnd++;
}
nNum[nP] = 0;
da(nNum, nP + 1, nEnd);
calHeight(nNum, nP);
int nLeft = 1, nRight = strlen(szStr), nMid;
int nAns = 0;
while (nLeft <= nRight)
{
nMid = (nLeft + nRight) / 2;
if (Check(nMid, nP, nN))
{
nLeft = nMid + 1;
nAns = nMid;
}
else nRight = nMid - 1;
}
printf("%d\n", nAns);
}
return 0;
}