题目链接,同步赛的时候不会 SA , 所以当时不会做这个题2333
如果我们想要求出长度为 r 的相同子串的对数
将满足 height[i]≥r 的 i 的集合与 i−1 的集合合并
那么每个集合内的子串都是 “ r 相似” 的
cnt[r] 表示 height[i]=r 时的方案数
max[r] 表示 height[i]=r 时的最大值
那么,按 height 值从大到小将集合合并,
合并时, 记 x ,y 为要合并的两个集合,那么
cnt[r] += size[x]*size[y]
ans[r] = max{max[x]*max[y] , min[x]*min[y]}
最后从大到小计算答案
cnt ′ [r] += cnt′[r + 1] + cnt[r]
ans ′ [r] = max{ans′[r + 1],ans[r]}
时间复杂度: O(N∗logN)
// savour uoj#131
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
#include <vector>
#include <utility>
#include <stack>
#include <queue>
#include <iostream>
#include <algorithm>
template<class Num>void read(Num &x)
{
char c; int flag = ;
while((c = getchar()) < '0' || c > '9')
if(c == '-') flag *= -;
x = c - '0';
while((c = getchar()) >= '0' && c <= '9')
x = (x<<) + (x<<) + (c-'0');
x *= flag;
return;
}
template<class Num>void write(Num x)
{
if(x < ) putchar('-'), x = -x;
static char s[];int sl = ;
while(x) s[sl++] = x% + '0',x /= ;
if(!sl) {putchar('0');return;}
while(sl) putchar(s[--sl]);
}
const int maxn = + ;
char s[maxn];
int a[maxn];
int n, sa[maxn], rank[maxn];
int c[maxn], height[maxn];
int fa[maxn], size[maxn];
long long max[maxn], min[maxn];
long long cnt, ans;
long long outc[maxn], outa[maxn];
struct Edge
{
int v, next;
Edge(int v = ,int next = ):v(v), next(next){}
}edge[maxn];
int head[maxn], el;
int find(int x)
{
return x == fa[x] ? x : (fa[x] = find(fa[x]));
}
void newedge(int u,int v)
{
edge[++el] = Edge(v, head[u]), head[u] = el;
}
void build_sa(int m)
{
static int t0[maxn], t1[maxn];
int *x = t0, *y = t1;
for(int i = ; i <= m; i++) c[i] = ;
for(int i = ; i <= n; i++) c[x[i] = s[i]]++;
for(int i = ; i <= m; i++) c[i] += c[i - ];
for(int i = n; i >= ; i--) sa[c[x[i]]--] = i;
for(int k = ; k <= n; k <<= )
{
int p = ;
for(int i = ; i < k; i++) y[++p] = n - i;
for(int i = ; i <= n; i++)
if(sa[i] > k) y[++p] = sa[i] - k;
for(int i = ; i <= m; i++) c[i] = ;
for(int i = ; i <= n; i++) c[x[y[i]]]++;
for(int i = ; i <= m; i++) c[i] += c[i - ];
for(int i = n; i >= ; i--) sa[c[x[y[i]]]--] = y[i];
std::swap(x, y);
x[sa[p = ]] = ;
for(int i = ; i <= n; i++)
x[sa[i]] = y[sa[i]] == y[sa[i - ]] && sa[i] + k <= n && sa[i - ] + k <= n && y[sa[i] + k] == y[sa[i - ] + k] ? p : ++p;
if(p == n) break;
m = p;
}
}
void build_height()
{
int k = ;
for(int i = ; i <= n; i++) rank[sa[i]] = i;
for(int i = ; i <= n; i++)
{
if(k != ) k--;
if(rank[i] == ) continue;
int j = sa[rank[i] - ];
while(s[j + k] == s[i + k]) k++;
height[rank[i]] = k;
}
}
void gather(int x,int y)
{
x = find(x), y = find(y);
if(x == y) return;
if(x > y) std::swap(x, y);
long long calc = std::max(max[x] * max[y], min[x] * min[y]);
if(!cnt || calc > ans) ans = calc;
cnt += (long long)size[x] * size[y];
fa[y] = x, size[x] += size[y];
max[x] = std::max(max[y], max[x]);
min[x] = std::min(min[y], min[x]);
}
void prework()
{
build_sa(), build_height();
for(int i = ; i <= n; i++)
newedge(height[i], i);
for(int i = ; i <= n; i++)
{
fa[i] = i, size[i] = ;
min[i] = max[i] = a[sa[i]];
}
}
void solve()
{
for(int i = n - ; i >= ; i--)
{
for(int j = head[i]; j ; j = edge[j].next)
gather(edge[j].v, edge[j].v - );
outc[i] = cnt, outa[i] = ans;
}
for(int i = ; i < n; i++)
{
write(outc[i]);
putchar(' ');
write(outa[i]), puts("");
}
}
void init()
{
read(n);
scanf("%s", s + );
for(int i = ; i <= n; i++) read(a[i]);
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("savour.in","r",stdin);
freopen("savour.out","w",stdout);
#endif
init();
prework();
solve();
#ifndef ONLINE_JUDGE
fclose(stdin);
fclose(stdout);
#endif
return ;
}