POJ 2774 Long Long Message （后缀数组）
The little cat is majoring in physics in the capital of Byterland. A piece of sad news comes to him these days: his mother is getting ill. Being worried about spending so much on railway tickets (Byterland is such a big country, and he has to spend 16 shours on train to his hometown), he decided only to send SMS with his mother.
The little cat lives in an unrich family, so he frequently comes to the mobile service center, to check how much money he has spent on SMS. Yesterday, the computer of service center was broken, and printed two very long messages. The brilliant little cat soon found out:
1. All characters in messages are lowercase Latin letters, without punctuations and spaces.
2. All SMS has been appended to each other – (i+1)-th SMS comes directly after the i-th one – that is why those two messages are quite long.
3. His own SMS has been appended together, but possibly a great many redundancy characters appear leftwards and rightwards due to the broken computer.
E.g: if his SMS is “motheriloveyou”, either long message printed by that machine, would possibly be one of “hahamotheriloveyou”, “motheriloveyoureally”, “motheriloveyouornot”, “bbbmotheriloveyouaaa”, etc.
4. For these broken issues, the little cat has printed his original text twice (so there appears two very long messages). Even though the original text remains the same in two printed messages, the redundancy characters on both sides would be possibly different.
You are given those two very long messages, and you have to output the length of the longest possible original text written by the little cat.
Background:
The SMS in Byterland mobile service are charging in dollars-per-byte. That is why the little cat is worrying about how long could the longest original text be.
Why ask you to write a program? There are four resions:
1. The little cat is so busy these days with physics lessons;
2. The little cat wants to keep what he said to his mother seceret;
3. POJ is such a great Online Judge;
4. The little cat wants to earn some money from POJ, and try to persuade his mother to see the doctor :(
Input
Output
Sample Input
yeshowmuchiloveyoumydearmotherreallyicannotbelieveit yeaphowmuchiloveyoumydearmother
Sample Output
27
分析：
题意是找两个字符串的最长公共子串的长度
可以用后缀数组来做。
先将两个字符串拼接成一个，中间用一个符号标记一下代表相隔两个字符串。
这就问题就转化成了求一个字符串中 出现两次的子串，并保证一个子串在标记之前，另一个在标记之后。
代码如下：
#include <cstdio> #include <iostream> #include <algorithm> #include <cstring> typedef long long ll; using namespace std; const int MAXN=200010; int wa[MAXN],wb[MAXN],wv[MAXN],Ws[MAXN]; int cmp(int *r,int a,int b,int l) {return r[a]==r[b]&&r[a+l]==r[b+l];} void da(const char r[],int sa[],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]=r[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++; } return; } int sa[MAXN],Rank[MAXN],height[MAXN]; void calheight(const char *r,int *sa,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) for(k?k--:0,j=sa[Rank[i]-1]; r[i+k]==r[j+k]; k++); for(int i=n;i>=1;--i) ++sa[i],Rank[i]=Rank[i-1]; } int main() { char r1[MAXN]; char r2[MAXN]; while(scanf("%s%s",r1,r2)!=EOF) { int L1=strlen(r1); int L2=strlen(r2); r1[L1]=126; int h=L1+1; for(int i=0;i<L2;i++) { r1[h++]=r2[i]; } r1[h]='\0'; int m=127; int len=h; da(r1,sa,h+1,m); calheight(r1,sa,h); int ans=0; for(int i=2;i<=len;i++) { if(height[i]>ans) { if(0<=sa[i-1]&&sa[i-1]<=L1&&L1<sa[i]) ans=height[i]; if(0<=sa[i]&&sa[i]<=L1&&L1<sa[i-1]) ans=height[i]; } } cout<<ans<<endl; } }
自己需要注意的问题： 字符串拼接之后 要在后面加'\0'
