Bzoj3238: [Ahoi2013]差异
2018-02-22 17:07
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题面
BzojSol
刚完品酒大会那道题后再看这道题发现这就是道\(SB\)题后缀数组+并查集
按\(height\)从大到小做
\(height\)是两个相邻\(rank\)的后缀的\(LCP\)
从大到小,那么每次合并\(height\)的两边的集合,同时记录答案
两边集合两两配对的\(LCP\)一定就是这个\(height\),乘法原理就可以了
这也算是一种套路吧
代码我常数大我最菜
# include <bits/stdc++.h> # define RG register # define IL inline # define Fill(a, b) memset(a, b, sizeof(a)) using namespace std; typedef long long ll; const int _(5e5 + 5); IL int Input(){ RG int x = 0, z = 1; RG char c = getchar(); 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, s[_], sa[_], rk[_], tmp[_], t[_], height[_], id[_]; int fa[_], size[_]; ll ans; char ss[_]; IL int Find(RG int x){ return x == fa[x] ? x : fa[x] = Find(fa[x]); } IL ll S(RG ll x){ return x * (x + 1) / 2; } IL int Cmp(RG int i, RG int j, RG int k){ return tmp[i] == tmp[j] && i + k <= n && j + k <= n && tmp[i + k] == tmp[j + k]; } IL void Suffix_Sort(){ RG int m = 26; for(RG int i = 1; i <= n; ++i) ++t[rk[i] = s[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) tmp[++l] = i; for(RG int i = 1; i <= n; ++i) if(sa[i] > k) tmp[++l] = sa[i] - k; for(RG int i = 1; i <= m; ++i) t[i] = 0; for(RG int i = 1; i <= n; ++i) ++t[rk[tmp[i]]]; for(RG int i = 1; i <= m; ++i) t[i] += t[i - 1]; for(RG int i = n; i; --i) sa[t[rk[tmp[i]]]--] = tmp[i]; swap(rk, tmp), 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, h = 0; i <= n; ++i){ if(h) --h; while(s[i + h] == s[sa[rk[i] - 1] + h]) ++h; height[rk[i]] = h; } } IL int _Cmp(RG int x, RG int y){ return height[x] > height[y]; } int main(RG int argc, RG char* argv[]){ scanf(" %s", ss + 1), n = strlen(ss + 1); for(RG int i = 1; i <= n; ++i){ s[i] = ss[i] - 'a' + 1, id[i] = fa[i] = i, size[i] = 1; ans += S(n - i) + 1LL * (n - i) * (n - i + 1); } Suffix_Sort(), sort(id + 2, id + n + 1, _Cmp); for(RG int i = 2; i <= n; ++i){ RG int x = Find(sa[id[i] - 1]), y = Find(sa[id[i]]); ans -= 2LL * size[x] * size[y] * height[id[i]]; fa[x] = y, size[y] += size[x]; } printf("%lld\n", ans); return 0; }
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