POJ-3261 Milk Patterns (后缀数组 最长k次重复子串)

题目：

http://poj.org/problem?id=3261

题意：

求n天牛奶质量序列中的最长k次重复子串，子串之间可重复。

思路：

标准后缀数组的模版题，利用后缀数组的height数组可以判断排名连续的k个后缀子串的前缀相同，则满足条件。

因为序列最多有2000个，所以只要对于整个序列离散化之后求后缀数组。然后二分枚举子串长度res，通过height数组扫一遍判断是否存在大于等于k个后缀子串的前缀有大于res个长度相同即可。

代码：

#include <bits/stdc++.h>
#define N 20005
using namespace std;

int n,m,sum,res,flag,k;
int seq
, sa
, ranks
, height
;
int wwa
, wwb
, wws
, wwv
;
bool cmp(int r[], int a, int b, int l)
{
return r[a] == r[b] && r[a+l] == r[b+l];
}
void da(int r[], int sa[], int n, int m)
{
int i, j, p, *x = wwa, *y = wwb;
for (i = 0; i < m; ++i) wws[i] = 0;
for (i = 0; i < n; ++i) wws[x[i]=r[i]]++;
for (i = 1; i < m; ++i) wws[i] += wws[i-1];
for (i = n-1; i >= 0; --i) sa[--wws[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) wwv[i] = x[y[i]];
for (i = 0; i < m; ++i) wws[i] = 0;
for (i = 0; i < n; ++i) wws[wwv[i]]++;
for (i = 1; i < m; ++i) wws[i] += wws[i-1];
for (i = n-1; i >= 0; --i) sa[--wws[wwv[i]]] = y[i];
for (swap(x, y), 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 calheight(int r[], int sa[], int n)
{
int i, j, k = 0;
for (i = 1; i <= n; ++i) ranks[sa[i]] = i;
for (i = 0; i < n; height[ranks[i++]] = k)
for (k?k--:0, j = sa[ranks[i]-1]; r[i+k] == r[j+k]; k++);
}
void solve()
{
//初始序0~n，最大值小于m。
seq
=0;
da(seq,sa,n+1,m);
calheight(seq,sa,n);
}
int temp
;
map<int,int>mp;
int main()
{
int i,j,kk,cas,T,t,x,y,z;
while(scanf("%d%d",&n,&k)!=EOF)
{
mp.clear();
for(i=0;i<n;i++)
{
scanf("%d",&seq[i]);
temp[i]=seq[i];
}
sort(temp,temp+n);
m=unique(temp,temp+n)-temp;
for(i=0;i<m;i++)
mp[temp[i]]=i+1;
for(i=0;i<n;i++)
seq[i]=mp[seq[i]];
solve();
x=1;y=n;
while(x<=y)
{
z=(x+y)/2;
flag=false;
for(i=2;i<=n&&!flag;i++)
{
t=1;
while(height[i]>=z)i++,t++;
if(t>=k)flag=true;
}
if(flag)
x=z+1,res=z;
else
y=z-1;
}
printf("%d\n",res);
}
return 0;
}