# 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

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