vjudge

一句话题意

给两个串，求最长公共子串。

sol

把两个串接在一起求后缀数组。其实中间最好用一个没有出现过的字符连接起来。

判断如果\(SA[i]\)和\(SA[i-1]\)不属于同一个串的话就可以拿\(Height[i]\)更新答案

code

#include<cstdio>

#include<algorithm>

#include<cstring>

#include<queue>

using namespace std;

#define ll long long

int gi()

{

	int x=0,w=1;char ch=getchar();

	while ((ch<'0'||ch>'9')&&ch!='-') ch=getchar();

	if (ch=='-') w=0,ch=getchar();

	while (ch>='0'&&ch<='9') x=(x<<3)+(x<<1)+ch-'0',ch=getchar();

	return w?x:-x;

}

const int N = 2e5+5;

char s1[N],s2[N];

int n,n1,n2,a[N],t[N],x[N],y[N],SA[N],Rank[N],Height[N],ans;

bool cmp(int i,int j,int k){return y[i]==y[j]&&y[i+k]==y[j+k];}

void getSA()

{

	int m=30;

	for (int i=1;i<=n;++i) ++t[x[i]=a[i]];

	for (int i=1;i<=m;++i) t[i]+=t[i-1];

	for (int i=n;i>=1;--i) SA[t[x[i]]--]=i;

	for (int k=1;k<=n;k<<=1)

	{

		int p=0;

		for (int i=0;i<=m;++i) y[i]=0;

		for (int i=n-k+1;i<=n;++i) y[++p]=i;

		for (int i=1;i<=n;++i) if (SA[i]>k) y[++p]=SA[i]-k;

		for (int i=0;i<=m;++i) t[i]=0;

		for (int i=1;i<=n;++i) ++t[x[y[i]]];

		for (int i=1;i<=m;++i) t[i]+=t[i-1];

		for (int i=n;i>=1;--i) SA[t[x[y[i]]]--]=y[i];

		swap(x,y);

		x[SA[1]]=p=1;

		for (int i=2;i<=n;++i) x[SA[i]]=cmp(SA[i],SA[i-1],k)?p:++p;

		if (p>=n) break;

		m=p;

	}

	for (int i=1;i<=n;++i) Rank[SA[i]]=i;

	for (int i=1,j=0;i<=n;++i)

	{

		if (j) --j;

		while (a[i+j]==a[SA[Rank[i]-1]+j]) ++j;

		Height[Rank[i]]=j;

	}

}

bool diff(int i,int j){return (i<=n1&&j>n1)||(i>n1&&j<=n1);}

int main()

{

	scanf("%s",s1+1);n1=strlen(s1+1);

	scanf("%s",s2+1);n2=strlen(s2+1);

	n=n1+n2;

	for (int i=1;i<=n1;++i) a[i]=s1[i]-96;

	for (int i=1;i<=n2;++i) a[n1+i]=s2[i]-96;

	getSA();

	for (int i=1;i<=n;++i)

		if (diff(SA[i],SA[i-1]))

			ans=max(ans,Height[i]);

	printf("%d\n",ans);

	return 0;

}