【POJ】3261 Milk Patterns

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

题意：一个长度为n的串，要求最长的子串的长度且这个子串的出现次数不少于k次。（1<=n<=20000, 2<=k<=n）

#include <cstdio>
#include <algorithm>
using namespace std;

const int N=20015;
void sort(int *x, int *y, int *sa, int n, int m) {
	static int c[N], i;
	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];
}
void hz(int *r, int *sa, int n, int m) {
	static int t1[N], t2[N];
	static int *x, *y, *t, j, i, p=0;
	x=t1; y=t2;
	for(i=0; i<n; ++i) x[i]=r[i], y[i]=i;
	sort(x, y, sa, n, m);
	for(j=1, p=1; p<n; j<<=1, m=p) {
		p=0;
		for(i=n-j; i<n; ++i) y[p++]=i;
		for(i=0; i<n; ++i) if(sa[i]-j>=0) y[p++]=sa[i]-j;
		sort(x, y, sa, n, m);
		for(t=x, x=y, y=t, x[sa[0]]=0, p=1, i=1; i<n; ++i)
			x[sa[i]]=y[sa[i]]==y[sa[i-1]]&&y[sa[i]+j]==y[sa[i-1]+j]?p-1:p++;
	}
}
void geth(int *a, int *sa, int *rank, int *h, int n) {
	static int k, i, j; k=0;
	for(i=1; i<=n; ++i) rank[sa[i]]=i;
	for(i=1; i<=n; h[rank[i++]]=k)
		for(k?--k:0, j=sa[rank[i]-1]; a[i+k]==a[j+k]; ++k);
}
const int oo=~0u>>2;
int sa[N], rank[N], h[N], n, a[N], b[N], K;
bool check(int k) {
	int cnt=1;
	for(int i=2; i<=n; ++i) {
		if(h[i]>=k) {
			++cnt;
			if(cnt>=K) return 1;
		}
		else cnt=1;
	}
	if(cnt>=K) return 1;
	return 0;
}
int mp[1000005];
int main() {
	scanf("%d%d", &n, &K);
	for(int i=1; i<=n; ++i) scanf("%d", &a[i]), b[i]=a[i];
	sort(b+1, b+1+n);
	int tot=unique(b+1, b+1+n)-b-1;
	for(int i=1; i<=tot; ++i) mp[b[i]]=i;
	for(int i=1; i<=n; ++i) a[i]=mp[a[i]];
	hz(a, sa, n+1, 200);
	geth(a, sa, rank, h, n);
	int mid, l=0, r=n;
	while(l<=r) {
		mid=(l+r)>>1;
		if(check(mid)) l=mid+1;
		else r=mid-1;
	}
	printf("%d\n", l-1);
	return 0;
}

　　

---

经典题...同样是分组height...将高度>=二分值的分在一组，然后判断是否有大于等于K个元素即可