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;
}