POJ 1734（后缀数组）

(http://acm.hust.edu.cn/vjudge/contest/view.action?cid=105595#problem/C)
题意：寻找最长的子串，在s中出现至少两次，子串所有字符可以全部+n/-n。
解法：这种问题统一解法，构造T[i] = S[i] - S[i-1]。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <queue>
//#include <tr1/unordered_set>
//#include <tr1/unordered_map>
#include <bitset>
//#pragma comment(linker, "/STACK:1024000000,1024000000")

using namespace std;

#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1
#define inf 1e9
#define debug(a) cout << #a" = " << (a) << endl;
#define debugarry(a, n) for (int i = 0; i < (n); i++) { cout << #a"[" << i << "] = " << (a)[i] << endl; }
#define clr(x, y) memset(x, y, sizeof x)
#define ll long long
#define ull unsigned long long
#define FOR(i,a,b) \
    for(i=a;a<b?i<=b:i>=b;a<b?i++:i--)

const int maxn = 20000*2+100;

struct suffix_array{
    int s[maxn];
    int sa[maxn],t[maxn],t2[maxn],c[maxn];
    int n;
    void build_sa(int m)
    {
        int i,*x=t,*y=t2;
        FOR(i,0,m-1) c[i]=0;
        FOR(i,0,n-1) c[x[i]=s[i]]++;
        FOR(i,1,m-1) c[i]+=c[i-1];
        FOR(i,n-1,0) sa[--c[x[i]]]=i;
        for(int k=1;k<=n;k<<=1)
        {
            int p=0;
            FOR(i,n-k,n-1) y[p++]=i;
            FOR(i,0,n-1) if(sa[i]>=k) y[p++]=sa[i]-k;

            FOR(i,0,m-1) c[i]=0;
            FOR(i,0,n-1) c[x[y[i]]]++;

            FOR(i,0,m-1) c[i]+=c[i-1];
            FOR(i,n-1,0) sa[--c[x[y[i]]]]=y[i];

            swap(x,y);
            p=1;
            x[sa[0]]=0;
            FOR(i,1,n-1)
                x[sa[i]]=y[sa[i-1]]==y[sa[i]]&&
                y[sa[i-1]+k]==y[sa[i]+k]?p-1:p++;
            if(p>=n) break;
            m=p;
        }
    }
    int Rank[maxn],height[maxn];
    void getHeight()
    {
        int k=0;
        for(int i=0;i<n;i++) Rank[sa[i]]=i;
        for(int i=0;i<n;i++){
            if(k)k--;
            int j=sa[Rank[i]-1];
            while(s[i+k]==s[j+k]) k++;
            height[Rank[i]]=k;
        }
    }
}su;

int num[maxn],n;

int cal(int len)
{
    for(int i=1;i<=n;i++)
    {
        int j;
        int minp=inf,maxp=-1;
        for(j=i;j<=n;j++)
        {
            minp=min(minp,su.sa[j]);
            maxp=max(maxp,su.sa[j]);
            if( su.height[j+1] < len )
                break;
        }
        if( maxp - minp > len )
            return true;
        i=j;
        //printf("i %d j %d\n",i,j);
    }
    return false;
}

int solve(){
    su.n=n+1;
    for(int i=0;i<n;i++)
        su.s[i]=num[i];
    su.build_sa(200);
    su.getHeight();
    int l=1,r=n,ans=0;
    while(l<=r)
    {
        int m=l+r>>1;
        if(cal(m)){
            ans=max(ans,m);
            l=m+1;
        }
        else r=m-1;
    }
    return ans;
}

int main()
{
//  freopen("input.txt","r",stdin);
//  freopen("output.txt","w",stdout);
    while(~scanf("%d",&n)){
        if(!n) break;
        for(int i=0;i<n;i++)
            scanf("%d",&num[i]);
        if( n < 10 )
        {
            printf("0\n");
            continue;
        }
        n--;
        for(int i=0;i<n;i++)
            num[i]=num[i]-num[i+1]+88;
        int ans=solve();
        if( ans < 4 )
            printf("0\n");
        else
            printf("%d\n",ans+1);
    }
    return 0;
}