poj 3261 Milk Patterns 后缀数组 最长重复子串
给一串数组，数组最少含有k个相同子串，可重叠，求这样子串的最长长度。
后缀数组求出 height[]，若连续k个height[]都大于mid，就可以了。当然要找最大的mid，这里可用二分查找.。
#include <iostream>
#include<stdio.h>
#include<stdlib.h>
using namespace std;
const int maxn = 1000005;
const int maxm = 1000001;
int wa[maxn], wb[maxn];
int r[maxn], b[1000001];
int rank[maxn];
int num[maxn], sa[maxn];
int height[200001];
int cmp (int *r, int a, int b, int l) {
return r[a] == r[b] && r[a+l] == r[b+l];
}
void make_sa(int n, int m){
int i, j, p, *x = wa, *y = wb;
for (i = 0; i < m; i++) b[i] = 0;
for (i = 0; i < n; i++) b[x[i] = num[i]]++;
for (i = 1; i < m; i++) b[i] += b[i-1];
for (i = n-1; i >= 0; i--) sa[--b[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++) r[i] = x[y[i]];
for (i = 0; i < m; i++) b[i] = 0;
for (i = 0; i < n; i++) b[r[i]]++;
for (i = 1; i < m; i++) b[i] += b[i-1];
for (i = n-1; i>= 0; i--) sa[--b[r[i]]] = y[i];
int *t = x; x = y; y = t;
for (i = p = 1, x[sa[0]] = 0; i < n; i++)
x[sa[i]] = cmp(y, sa[i-1], sa[i], j) ? p-1:p++;
}
return;
}
void calheight(int len)
{
int i,j,k=0;
for (i =1; i<len+1; i++) rank[sa[i]] = i;
for(i=0;i<len;height[rank[i++]]=k)
for(k?k--:0,j=sa[rank[i]-1];num[i+k]==num[j+k];k++);
return ;
}
int k;
bool judge(int key,int len)
{
int ssum=0;
for(int i=1;i<len;i++)
{
if(height[i]>=key)
{
ssum++;
if(ssum+1>=k)
{
return true;
}
}
else{
ssum=0;
}
}
return false;
}
int main()
{
int n,i;
while(scanf("%d%d",&n,&k)!=EOF)
{
for(i=0;i<n;i++)
{
scanf("%d",&num[i]);
}
num[n]=0;
make_sa(n+1,maxm);
calheight(n);
int low=1,hight=n;
while(low<=hight)
{
int mid=(low+hight)>>1;
if(judge(mid,n+1))
{
low=mid+1;
}
else{
hight=mid-1;
}
}
if(hight<n)
{
printf("%d\n",hight);
}
}
return 0;
}