HDU - 1403 后缀数组初步

题意:求两个串的最长公共子串

两个串连接起来然后求高度数组

注意两个sa值必须分别在不同一侧

本题是用来测试模板的,回想起青岛那次翻车感觉很糟糕

#include<iostream>

#include<algorithm>

#include<cstdio>

#include<cstring>

#include<cstdlib>

#include<cmath>

#include<string>

#include<vector>

#include<stack>

#include<queue>

#include<set>

#include<map>

#define rep(i,j,k) for(register int i=j;i<=k;i++)

#define rrep(i,j,k) for(register int i=j;i>=k;i--)

#define erep(i,u) for(register int i=head[u];~i;i=nxt[i])

#define iin(a) scanf("%d",&a)

#define lin(a) scanf("%lld",&a)

#define din(a) scanf("%lf",&a)

#define s0(a) scanf("%s",a)

#define s1(a) scanf("%s",a+1)

#define print(a) printf("%lld",(ll)a)

#define enter putchar('\n')

#define blank putchar(' ')

#define println(a) printf("%lld\n",(ll)a)

#define IOS ios::sync_with_stdio(0)

using namespace std;

const int maxn = 1e6+11;

const int oo = 0x3f3f3f3f;

const double eps = 1e-7;

typedef long long ll;

ll read(){

    ll x=0,f=1;register char ch=getchar();

    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}

    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}

    return x*f;

}

char str[maxn];

struct SA{

	int Rank[maxn],sa[maxn],tsa[maxn],A[maxn],B[maxn];

	int cntA[maxn],cntB[maxn];

	int height[maxn],best[maxn][30],n;//height[i]:第sa[i]与sa[i-1]的cp

	void get(){

		n=strlen(str+1);

		rep(i,0,300) cntA[i]=0;

		rep(i,1,n) cntA[str[i]]++;

		rep(i,1,300) cntA[i]+=cntA[i-1];

		rrep(i,n,1) sa[cntA[str[i]]--]=i;

		Rank[sa[1]]=1;

		rep(i,2,n){

			if(str[sa[i]]==str[sa[i-1]]){

				Rank[sa[i]]=Rank[sa[i-1]];

			}else{

				Rank[sa[i]]=1+Rank[sa[i-1]];

			}

		}

		for(int l=1;Rank[sa[n]]<n;l<<=1){

			rep(i,1,n) cntA[i]=cntB[i]=0;

			rep(i,1,n) cntA[A[i]=Rank[i]]++;

			rep(i,1,n) cntB[B[i]=(i+l<=n?Rank[i+l]:0)]++;

			rep(i,1,n) cntA[i]+=cntA[i-1],cntB[i]+=cntB[i-1];

			rrep(i,n,1) tsa[cntB[B[i]]--]=i;

			rrep(i,n,1) sa[cntA[A[tsa[i]]]--]=tsa[i];

			Rank[sa[1]]=1;

			rep(i,2,n){

				bool flag=A[sa[i]]==A[sa[i-1]]&&B[sa[i]]==B[sa[i-1]];

				flag=!flag;

				Rank[sa[i]]=Rank[sa[i-1]]+flag;

			}

		}

	}

	void ht(){

		int j=0;

		rep(i,1,n){

			if(j) j--;

			while(str[i+j]==str[sa[Rank[i]-1]+j]) j++;

			height[Rank[i]]=j;

		}

	}

	void rmq(){

		rep(i,1,n) best[i][0]=height[i];

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

			for(int j=1;j+(1<<i)-1<=n;j++){

				best[j][i]=min(best[j][i-1],best[j+(1<<(i-1))][i-1]);

			}

		}

	}

	int query(int l,int r){

		if(l==r)return -oo;

		if(l>r)swap(l,r);

		l++;

		int k=log2(r-l+1);

		return min(best[l][k],best[r-(1<<k)+1][k]);

	}

}sa;

char str1[maxn];

int main(){

	while(~s1(str)){

		int len=strlen(str+1);

		s1(str1);

		strcat(str+1,str1+1);

		sa.get();

		sa.ht();

		int n=strlen(str+1);

		int ans=0;

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

			if(ans<sa.height[i]){

				if((sa.sa[i]<=len&&sa.sa[i-1]>len)||(sa.sa[i]>len&&sa.sa[i-1]<=len)){

					ans=sa.height[i];

				}

			}

		}

		println(ans);

	}

	return 0;

}

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巴特西

HDU - 1403 后缀数组初步

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