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求出后缀数组后, 对height排序, 从大到小来处理(r相似必定是0~r-1相似), 并查集维护. 复杂度O(NlogN + Nalpha(N))
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#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const ll INF = -1LL << 60;
const int maxn = 300009;
char S[maxn];
int w[maxn], N;
int cnt[maxn], Sa[maxn], Rank[maxn], Height[maxn];
int mx[maxn], mn[maxn], fa[maxn], sz[maxn], r[maxn];
ll ans[maxn], tot[maxn];
void Init() {
scanf("%d%s", &N, S);
for(int i = 0; i < N; i++) scanf("%d", w + i);
S[N++] = ‘$‘;
}
void Build() {
int m = 255, *x = Height, *y = Rank;
for(int i = 0; i < m; i++) cnt[i] = 0;
for(int i = 0; i < N; i++) cnt[x[i] = S[i]]++;
for(int i = 1; i < m; i++) cnt[i] += cnt[i - 1];
for(int i = N; i--; ) Sa[--cnt[x[i]]] = i;
for(int k = 1, p = 0; k <= N; k <<= 1, p = 0) {
for(int i = N - k; i < N; i++) y[p++] = i;
for(int i = 0; i < N; i++)
if(Sa[i] >= k) y[p++] = Sa[i] - k;
for(int i = 0; i < m; i++) cnt[i] = 0;
for(int i = 0; i < N; i++) cnt[x[y[i]]]++;
for(int i = 1; i < m; i++) cnt[i] += cnt[i - 1];
for(int i = N; i--; ) Sa[--cnt[x[y[i]]]] = y[i];
swap(x, y), x[Sa[0]] = 0, p = 1;
for(int i = 1; i < N; i++) {
if(y[Sa[i]] != y[Sa[i - 1]] || y[Sa[i] + k] != y[Sa[i - 1] + k]) p++;
x[Sa[i]] = p - 1;
}
if((m = p) >= N) break;
}
for(int i = 0; i < N; i++) Rank[Sa[i]] = i;
Height[0] = 0;
for(int i = 0, h = 0; i < N; i++) if(Rank[i]) {
if(h) h--;
while(S[i + h] == S[Sa[Rank[i] - 1] + h]) h++;
Height[Rank[i]] = h;
}
}
int Find(int x) {
return x == fa[x] ? x : fa[x] = Find(fa[x]);
}
bool Cmp(const int &l, const int &r) {
return Height[l] > Height[r];
}
inline void Max(ll &x, ll t) {
if(t > x) x = t;
}
inline void Max(int &x, int t) {
if(t > x) x = t;
}
inline void Min(int &x, int t) {
if(t < x) x = t;
}
void Work() {
for(int i = 0; i < N; i++) {
sz[i] = 1;
r[i] = fa[i] = i;
mx[i] = mn[i] = w[i];
tot[i] = 0;
ans[i] = INF;
}
sort(r, r + N, Cmp);
for(int i = 0; i < N; i++) if(r[i] > 1) {
int u = Find(Sa[r[i]]), v = Find(Sa[r[i] - 1]);
if(u == v) continue;
tot[Height[r[i]]] += ll(sz[u]) * sz[v];
Max(ans[Height[r[i]]], max(ll(mx[u]) * mx[v], ll(mn[u]) * mn[v]));
fa[u] = v, sz[v] += sz[u];
Max(mx[v], mx[u]);
Min(mn[v], mn[u]);
}
for(int i = Height[r[0]]; i--; )
tot[i] += tot[i + 1], Max(ans[i], ans[i + 1]);
for(int i = 0; i + 1 < N; i++)
printf("%lld %lld\n", tot[i], ans[i] != INF ? ans[i] : 0);
}
int main() {
Init();
Build();
Work();
return 0;
}
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BZOJ 4199: [Noi2015]品酒大会( 后缀数组 + 并查集 )
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原文地址：http://www.cnblogs.com/JSZX11556/p/5156166.html
