POJ 3261 USACO 2006 December Gold Milk Patterns

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题目大意：给出一个字符串，求出出现过k次以上的最长的子串(可重叠).

思路：现弄出来sa数组和height数组，之后就是判断每个长度为k的height数组的区间中最小的数字的最大值了.为什么好多人都二分了?这只要单调队列扫一次就行了啊..

CODE:

#include <queue>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define MAX 1000010using namespace std;int len,k;int s[MAX],val[MAX],sa[MAX];int rank[MAX],height[MAX];inline bool Same(int x,int y,int l){return val[x] == val[y] && ((x + l >= len && y + l >= len) || (x + l < len && y + l < len && val[x + l] == val[y + l]));}void GetSuffixArray(){static int _val[MAX],cnt[MAX],q[MAX],lim = 1000001;for(int i = 0; i < len; ++i)++cnt[val[i] = s[i]];for(int i = 1; i < lim; ++i)cnt[i] += cnt[i - 1];for(int i = len - 1; ~i; --i)sa[--cnt[val[i]]] = i;for(int d = 1;; ++d) {int top = 0,l = 1 << (d - 1);for(int i = 0; i < len; ++i)if(sa[i] + l >= len)q[top++] = sa[i];for(int i = 0; i < len; ++i)if(sa[i] >= l)q[top++] = sa[i] - l;for(int i = 0; i < lim; ++i)cnt[i] = 0;for(int i = 0; i < len; ++i)++cnt[val[q[i]]];for(int i = 1; i < len; ++i)cnt[i] += cnt[i - 1];for(int i = len - 1; ~i; --i)sa[--cnt[val[q[i]]]] = q[i];lim = 0;for(int i = 0,j; i < len; ++lim) {for(j = i; j < len - 1 && Same(sa[j],sa[j + 1],l); ++j);for(; i <= j; ++i)_val[sa[i]] = lim;}for(int i = 0; i < len; ++i)val[i] = _val[i];if(lim == len)break;}return ;}void GetHeight(){for(int i = 0; i < len; ++i)rank[sa[i]] = i;for(int k = 0,i = 0; i < len; ++i) {if(k)--k;int j = sa[rank[i] - 1];while(s[i + k] == s[j + k])++k;height[rank[i]] = k;}}struct Complex{int pos,val;Complex(int _,int __):pos(_),val(__) {}Complex() {}};int main(){cin >> len >> k;for(int i = 0; i < len; ++i)scanf("%d",&s[i]);GetSuffixArray();GetHeight();deque<Complex> q;int ans = 0;for(int i = 0; i < len; ++i) {while(!q.empty() && height[i] <= q.back().val)q.pop_back();while(!q.empty() && i - q.front().pos >= k - 1)q.pop_front();q.push_back(Complex(i,height[i]));ans = max(ans,q.front().val);}cout << ans << endl;return 0;}

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