POJ 3261 Milk Patterns (后缀数组 + 二分)
2017-04-28 18:07
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题意:
给你一个两万的数组, 求可重叠的K次最长重复子串。
思路:
后缀数组经典问题
类似于求不可重叠的最长重复子串。
先求一遍后缀数组, 给sa数组 分块。
二分长度m
每一块 height 必须大于等于 长度m。
然后如果存在一块, 满足数量大于等于k 那么这个长度m就是合法的。 继续二分即可。
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#define Siz(x) (int)x.size()
using namespace std;
const int maxn = 1000000 + 10;
int t1[maxn], t2[maxn], c[maxn];
bool cmp(int* r, int a,int b,int l){
return r[a] == r && r[a+l] == r[b+l];
}
void da(int str[], int sa[], int Rank[], int lcp[], int n, int m){
++n;
int i, j, p, *x = t1, *y = t2;
for (i = 0; i < m; ++i) c[i] = 0;
// puts("hha");
for (i = 0; i < n; ++i) c[x[i] = str[i] ]++;
for (i = 1; i < m; ++i) c[i] += c[i-1];
for (i = n-1; i >= 0; --i) sa[--c[x[i] ] ] = i;
for (j = 1; j <= n; j <<= 1){
p = 0;
for (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 < m; ++i) c[i] = 0;
for (i = 0; i < n; ++i) c[x[y[i] ] ]++;
for (i = 1; i < m; ++i) c[i] += c[i-1];
for (i = n-1; i >= 0; --i) sa[--c[x[y[i] ] ] ] = y[i];
swap(x,y);
p = 1; x[sa[0] ] = 0;
for (i = 1; i < n; ++i){
x[sa[i] ] = cmp(y, sa[i-1], sa[i], j) ? p-1 : p++;
}
if (p >= n)break;
m= p;
}
int k = 0;
n--;
for (i = 0; i <= n; ++i) Rank[sa[i] ] = i;
for (i = 0; i < n; ++i){
if (k)--k;
j = sa[Rank[i]-1 ];
while(str[i+k] == str[j+k])++k;
lcp[Rank[i] ] = k;
}
}
int lcp[maxn], a[maxn], sa[maxn], Rank[maxn];
int n,k;
int bit[maxn];
bool judge(int m){
int cnt = 0;
memset(bit,0,sizeof bit);
bit[0]++;
for (int i = 1; i < n; ++i){
int id = i + 1;
if (lcp[id] >= m){
bit[cnt]++;
}
else bit[++cnt]++;
}
for (int i = 0; i <= cnt; ++i){
if (bit[i] >= k) return true;
}
return false;
}
int main(){
while(~scanf("%d %d",&n, &k)){
for (int i = 0; i < n; ++i) scanf("%d",a+i);
a
= 0;
da(a,sa,Rank,lcp,n,1000000);
int l = 1, r = n, m;
while(l <= r){
m = l + r >> 1;
if (judge(m)) l = m + 1;
else r = m-1;
}
printf("%d\n",r);
}
return 0;
}
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给你一个两万的数组, 求可重叠的K次最长重复子串。
思路:
后缀数组经典问题
类似于求不可重叠的最长重复子串。
先求一遍后缀数组, 给sa数组 分块。
二分长度m
每一块 height 必须大于等于 长度m。
然后如果存在一块, 满足数量大于等于k 那么这个长度m就是合法的。 继续二分即可。
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#define Siz(x) (int)x.size()
using namespace std;
const int maxn = 1000000 + 10;
int t1[maxn], t2[maxn], c[maxn];
bool cmp(int* r, int a,int b,int l){
return r[a] == r && r[a+l] == r[b+l];
}
void da(int str[], int sa[], int Rank[], int lcp[], int n, int m){
++n;
int i, j, p, *x = t1, *y = t2;
for (i = 0; i < m; ++i) c[i] = 0;
// puts("hha");
for (i = 0; i < n; ++i) c[x[i] = str[i] ]++;
for (i = 1; i < m; ++i) c[i] += c[i-1];
for (i = n-1; i >= 0; --i) sa[--c[x[i] ] ] = i;
for (j = 1; j <= n; j <<= 1){
p = 0;
for (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 < m; ++i) c[i] = 0;
for (i = 0; i < n; ++i) c[x[y[i] ] ]++;
for (i = 1; i < m; ++i) c[i] += c[i-1];
for (i = n-1; i >= 0; --i) sa[--c[x[y[i] ] ] ] = y[i];
swap(x,y);
p = 1; x[sa[0] ] = 0;
for (i = 1; i < n; ++i){
x[sa[i] ] = cmp(y, sa[i-1], sa[i], j) ? p-1 : p++;
}
if (p >= n)break;
m= p;
}
int k = 0;
n--;
for (i = 0; i <= n; ++i) Rank[sa[i] ] = i;
for (i = 0; i < n; ++i){
if (k)--k;
j = sa[Rank[i]-1 ];
while(str[i+k] == str[j+k])++k;
lcp[Rank[i] ] = k;
}
}
int lcp[maxn], a[maxn], sa[maxn], Rank[maxn];
int n,k;
int bit[maxn];
bool judge(int m){
int cnt = 0;
memset(bit,0,sizeof bit);
bit[0]++;
for (int i = 1; i < n; ++i){
int id = i + 1;
if (lcp[id] >= m){
bit[cnt]++;
}
else bit[++cnt]++;
}
for (int i = 0; i <= cnt; ++i){
if (bit[i] >= k) return true;
}
return false;
}
int main(){
while(~scanf("%d %d",&n, &k)){
for (int i = 0; i < n; ++i) scanf("%d",a+i);
a
= 0;
da(a,sa,Rank,lcp,n,1000000);
int l = 1, r = n, m;
while(l <= r){
m = l + r >> 1;
if (judge(m)) l = m + 1;
else r = m-1;
}
printf("%d\n",r);
}
return 0;
}
Milk Patterns
Farmer John has noticed that the quality of milk given by his cows varies from day to day. On further investigation, he discovered that although he can't predict the quality of milk from one day to the next, there are some regular patterns in the daily milk quality. To perform a rigorous study, he has invented a complex classification scheme by which each milk sample is recorded as an integer between 0 and 1,000,000 inclusive, and has recorded data from a single cow over N (1 ≤ N ≤ 20,000) days. He wishes to find the longest pattern of samples which repeats identically at least K (2 ≤ K ≤ N) times. This may include overlapping patterns -- 1 2 3 2 3 2 3 1 repeats 2 3 2 3 twice, for example. Help Farmer John by finding the longest repeating subsequence in the sequence of samples. It is guaranteed that at least one subsequence is repeated at least K times. Input Line 1: Two space-separated integers: N and K Lines 2..N+1: N integers, one per line, the quality of the milk on day i appears on the ith line. Output Line 1: One integer, the length of the longest pattern which occurs at least K times Sample Input 8 2 1 2 3 2 3 2 3 1 Sample Output 4 Source USACO 2006 December Gold |
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