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POJ 2774 Long Long Message ——后缀数组

【题目分析】

    用height数组RMQ的性质去求最长的公共子串。

    要求sa[i]和sa[i-1]必须在两个串中,然后取height的MAX。

    利用中间的字符来连接两个字符串的思想很巧妙,记得最后还需要空一个位置避免冲突。

【代码】

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>

#include <map>
#include <set>
#include <queue>
#include <string>
#include <iostream>
#include <algorithm>

using namespace std;

#define maxn 200005 
#define inf 0x3f3f3f3f
#define F(i,j,k) for (int i=j;i<=k;++i)
#define D(i,j,k) for (int i=j;i>=k;--i)

void Finout()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
//    freopen("out.txt","w",stdout);
    #endif
}

int Getint()
{
    int x=0,f=1; 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;
}

int tot=0,l1,l2;
char s1[maxn],s2[maxn];

struct Suf_Arr{
	int s[maxn<<1],l;
	int wa[maxn],wb[maxn],wv[maxn],ws[maxn];
	int sa[maxn],rank[maxn],height[maxn];
	int cmp(int *r,int a,int b,int l)
	{return r[a]==r[b]&&r[a+l]==r[b+l];}
	void getsa(int n,int m)
	{
		int i,j,p,*x=wa,*y=wb,*t;
		for (i=0;i<m;++i) ws[i]=0;
		for (i=0;i<n;++i) ws[x[i]=s[i]]++;
		for (i=1;i<m;++i) ws[i]+=ws[i-1];
		for (i=n-1;i>=0;--i) sa[--ws[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) ws[i]=0;
			for (i=0;i<n;++i) ws[wv[i]]++;
			for (i=1;i<m;++i) ws[i]+=ws[i-1];
			for (i=n-1;i>=0;--i) sa[--ws[wv[i]]]=y[i];
			for (t=x,x=y,y=t,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++;
		}
	}
	void gethi(int n)
	{
		int i,j,k=0;
		for (i=1;i<=n;++i) rank[sa[i]]=i;
		for (int i=0;i<n;height[rank[i++]]=k)
			for (k?k--:0,j=sa[rank[i]-1];s[i+k]==s[j+k];k++);
	}
	void build()
	{
		getsa(l+2,30);
//		for (int i=0;i<=l;++i) cout<<sa[i]<<" "; cout<<endl;
		gethi(l+1);
	}
	void solve()
	{
		int maxx=0;
		for(int i=2;i<l;++i)
			if (height[i]>maxx)
			{
				if (0<=sa[i-1]&&sa[i-1]<l1&&l1<sa[i]) maxx=height[i];
				if (0<=sa[i]&&sa[i]<l1&&l1<sa[i-1]) maxx=height[i];
			}
		printf("%d\n",maxx);
	}
}arr;

int main()
{
    Finout();
    scanf("%s",s1); scanf("%s",s2); 
    l1=strlen(s1); l2=strlen(s2); arr.l=l1+l2+1;
    for (int i=0;i<l1;++i) arr.s[i]=s1[i]-‘a‘+1;
    for (int i=0;i<l2;++i) arr.s[i+l1+1]=s2[i]-‘a‘+1;
    arr.s[l1+l2+1]=0; arr.s[l1]=28;
//    printf("len is %d\n",arr.l);
//    for (int i=0;i<=arr.l;++i) cout<<(char)(arr.s[i]+‘a‘-1); cout<<endl;
    arr.build();
    arr.solve();
}

  

POJ 2774 Long Long Message ——后缀数组