POJ3261：Milk Patterns

题面

vjudge

Sol

二分答案+分组，判断有没有一个组的后缀个数不小于 k
做法

# include <bits/stdc++.h>
# define IL inline
# define RG register
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(20010);

IL ll Read(){
	RG char c = getchar(); RG ll x = 0, z = 1;
	for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
	for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
	return x * z;
}

int n, a[_], sa[_], rk[_], y[_], h[_], height[_], t[1000010], ans, K, mx;

IL bool Cmp(RG int i, RG int j, RG int k){  return y[i] == y[j] && y[i + k] == y[j + k];  }

IL void Sort(){
	RG int m = mx;
	for(RG int i = 1; i <= n; ++i) ++t[rk[i] = a[i]];
	for(RG int i = 1; i <= m; ++i) t[i] += t[i - 1];
	for(RG int i = n; i; --i) sa[t[rk[i]]--] = i;
	for(RG int k = 1; k <= n; k <<= 1){
		RG int l = 0;
		for(RG int i = n - k + 1; i <= n; ++i) y[++l] = i;
		for(RG int i = 1; i <= n; ++i) if(sa[i] > k) y[++l] = sa[i] - k;
		for(RG int i = 0; i <= m; ++i) t[i] = 0;
		for(RG int i = 1; i <= n; ++i) ++t[rk[y[i]]];
		for(RG int i = 1; i <= m; ++i) t[i] += t[i - 1];
		for(RG int i = n; i; --i) sa[t[rk[y[i]]]--] = y[i];
		swap(rk, y); rk[sa[1]] = l = 1;
		for(RG int i = 2; i <= n; ++i) rk[sa[i]] = Cmp(sa[i - 1], sa[i], k) ? l : ++l;
		if(l >= n) break; m = l;
	}
	for(RG int i = 1; i <= n; ++i){
		h[i] = max(0, h[i - 1] - 1);
		if(rk[i] == 1) continue;
		while(a[i + h[i]] == a[sa[rk[i] - 1] + h[i]]) ++h[i];
	}
	for(RG int i = 1; i <= n; ++i) height[i] = h[sa[i]];
}

IL bool Check(RG int x){
	RG int cnt = 0;
	for(RG int i = 2; i <= n; ++i){
		if(height[i] < x) cnt = 0;
		else cnt++;
		if(cnt >= K - 1) return 1;
	}
	return 0;
}

int main(RG int argc, RG char* argv[]){
	n = Read(); K = Read();
	for(RG int i = 1; i <= n; ++i) a[i] = Read(), mx = max(a[i], mx);
	Sort();
	RG int l = 0, r = n;
	while(l <= r){
		RG int mid = (l + r) >> 1;
		if(Check(mid)) ans = mid, l = mid + 1;
		else r = mid - 1;
	}
	printf("%d\n", ans);
	return 0;
}