POJ 3882 Stammering Aliens 后缀数组height应用

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题目来源：POJ 3882 Stammering Aliens

题意：给你m一个一个字符串求至少出现m次的最长字符串可以在字符串中重叠出现

思路：二分长度l 然后从height数组中找长度大于等于l的前缀

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 40010;char s[maxn];int sa[maxn];int t[maxn], t2[maxn], c[maxn];int rank[maxn], height[maxn];int pos, lim;void build_sa(int m, int n){int i, *x = t, *y = t2;for(i = 0; i < m; i++)c[i] = 0;for(i = 0; i < n; i++)c[x[i] = s[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(int k = 1; k <= n; k <<= 1){int p = 0;for(i = n-k; i < n; i++)y[p++] = i;for(i = 0; i < n; i++)if(sa[i] >= k)y[p++] = sa[i] - k;for(i = 0; i < m; i++)c[i] = 0;for(i = 0; i < n; i++)c[x[y[i]]]++;for(i = 0; 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]] = y[sa[i-1]] == y[sa[i]] && y[sa[i-1]+k] == y[sa[i]+k] ? p-1 : p++;if(p >= n)break;m = p;}}void getHeight(int n){int k = 0;for(int i = 0; i <= n; i++)rank[sa[i]] = i;for(int i = 0; i < n; i++){if(k)k--;int j = sa[rank[i]-1];while(s[i+k] == s[j+k])k++;height[rank[i]] = k;}}bool ok(int l, int n){int flag = 0;int cnt = 1;pos = 0;int p = -1;for(int i = 1; i <= n; i++){if(height[i] >= l){cnt++;p = max(p, sa[i]);}else{p = sa[i];cnt = 1;}if(cnt >= lim){flag = 1;pos = max(pos, p);} }if(flag)return true;return false;}int main(){int l, r, n;while(scanf("%d", &lim) && lim){scanf("%s", s);n = strlen(s);if(lim == 1){printf("%d 0\n", n);continue;}l = 1;r = n;int ans = -1;int ans2;build_sa(256, n+1);getHeight(n);while(l <= r){int m = (l + r) >> 1;if(ok(m, n)){ans = m;ans2 = pos;l = m+1;}elser = m-1;}if(ans == -1)puts("none");else{printf("%d %d\n", ans, ans2);}}return 0;}

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