后缀数组 - poj1743 Musical Theme

题目：

http://poj.org/problem?id=1743

题意：

给一个序列，对序列中一段区间[l,r]，若此区间中的数均加上或减去同一个数后，与序列的另一区间[l1,r1]相同，且[l,r]与[l1,r1]不交，称这一区间重复出现，求序列中至少重复出现一次的最长区间长度，若该长度小于5输出0

思路：

看起来很麻烦，事实上做一个简单转化，两段区间重复即意味着这两段区间的数的差值相同，用序列的第i个数减去第i-1个数，得到一个新序列，问题转化为求新序列中的最长不可重叠重复子串的长度，后缀数组能解决的经典问题之一

对新序列建立后缀数组，二分枚举答案，问题转化为判定是否存在长度为mid的不重叠子串，将后缀序列视为若干组，要求每组内满足任意两个后缀的lcp值均大于等于mid，每个组即在height数组中对应一段连续区间[l,r]，检查rank值为[l-1,r]的后缀，判断这个分组中是否存在两个不重叠的后缀即可

代码：

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#define rep(i,n) for(int i = 0; i < n; i++)
using namespace std;

const int MAXSIZE = 2*1e4 + 100;
const int MAXINT = 0x71ffffff;

int rk[MAXSIZE], sa[MAXSIZE], height[MAXSIZE], wa[MAXSIZE], res[MAXSIZE];
int w[MAXSIZE];  //转储待处理字符串
int len;

void getSa(int up) {
int *k = rk, *id = height, *r = res, *cnt = wa;
rep(i, up) cnt[i] = 0;
rep(i, len) cnt[k[i] = w[i]]++;
rep(i, up) cnt[i + 1] += cnt[i];
for (int i = len - 1; i >= 0; i--) {
sa[--cnt[k[i]]] = i;
}
int d = 1, p = 0;
while (p < len){
for (int i = len - d; i < len; i++) id[p++] = i;
rep(i, len)  if (sa[i] >= d) id[p++] = sa[i] - d;
rep(i, len) r[i] = k[id[i]];
rep(i, up) cnt[i] = 0;
rep(i, len) cnt[r[i]]++;
rep(i, up) cnt[i + 1] += cnt[i];
for (int i = len - 1; i >= 0; i--) {
sa[--cnt[r[i]]] = id[i];
}
swap(k, r);
p = 0;
k[sa[0]] = p++;
rep(i, len - 1) {
if (sa[i] + d < len && sa[i + 1] + d < len && r[sa[i]] == r[sa[i + 1]] && r[sa[i] + d] == r[sa[i + 1] + d])
k[sa[i + 1]] = p - 1;
else k[sa[i + 1]] = p++;
}
if (p >= len) return;
d <<= 1, up = p, p = 0;
}
}

//计算rank及height值
void getHeight() {
int i, k, h = 0;
rep(i, len) rk[sa[i]] = i;
rep(i, len) {
if (rk[i] == 0)
h = 0;
else {
k = sa[rk[i] - 1];
if (h) h--;
while (w[i + h] == w[k + h]) h++;
}
height[rk[i]] = h;
}
}

void getSuffix() {
int up = 200;
len--;
w[len] = 0;
getSa(up + 1);
getHeight();
}

int s[MAXSIZE];

int main(){
cin>>len;
while (len!=0){
for (int i=0;i<len;++i){
scanf("%d",s+i);
}
for (int i = 0;i<len-1;++i)
w[i] = s[i+1] - s[i] + 88;
getSuffix();

int l=0,r=len;
while (l<=r){
int mid = (l+r)>>1;
//cout<<l<<" "<<r<<" "<<mid<<endl;
int mininum = sa[0], maxinum = sa[0];
bool flag = false;
for (int i=1;i<len;++i){
if (flag) break;
if (height[i]>=mid){
mininum = min(mininum, sa[i]);
maxinum = max(maxinum, sa[i]);
}
else {
if (maxinum - mininum > mid) flag  = true;
mininum = sa[i];
maxinum = sa[i];
}
}
if (maxinum - mininum > mid) flag  = true;
//cout<<mid<<" "<<flag<<endl;
if (flag)
l = mid + 1;
else
r = mid - 1;
}
if (r+1>=5) cout<<r+1<<endl;
else cout<<0<<endl;
cin>>len;
}
return 0;
}