[NOI2015]品酒大会

题面：

给出一个n个字符的字符串，每个字符有一个权值，现在要求求出有多少种方法可以选出两个长度为r的相同的子串，以及能选出来的两个串首字符的权值的最大乘积。(可能描述的有点shi)

(Solution:)

因为子串是后缀的前缀，所以就自然地想到用后缀树数组。

因为两个 (k) 相似的串必定满足 (r) 相似 ((lle k)).

所以我们离线从大到小处理 (Height) 数组, 这样满足的 (height) 区间只会变大，不会变小，把这些后缀想象成一个个集合，然后就可以用带权并查集维护了。每次处理一个 (height) ，那么它关联到的两个后缀的前缀是满足条件的，于是 (merge) 这两个后缀(实际上是集合)。

还有一个很重要的地方是每次处理一个 (height) 时都只算了height相似的答案，而这个答案是会对之后处理的答案做出全部贡献的，即两个 (k) 相似的串必定满足 (r) 相似 ((lle k)).所以再做一遍前缀和，最大乘积就取 (max)。

(ps:别忘了维护最小值，因为负负得正)

(Source:)

#include <set>
#include <cmath>
#include <cctype>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <assert.h>
#include <algorithm>

using namespace std;

#define fir first
#define sec second
#define pb push_back
#define mp make_pair
#define LL long long
#define INF (0x3f3f3f3f)
#define mem(a, b) memset(a, b, sizeof (a))
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define Debug(x) cout << #x << " = " << x << endl
#define tralve(i, x) for (register int i = head[x]; i; i = nxt[i])
#define For(i, a, b) for (register int (i) = (a); (i) <= (b); ++ (i))
#define Forr(i, a, b) for (register int (i) = (a); (i) >= (b); -- (i))
#define file(s) freopen(s".in", "r", stdin), freopen(s".out", "w", stdout)
#define ____ debug("go
")

namespace io {
    static char buf[1<<21], *pos = buf, *end = buf;
    inline char getc()
    { return pos == end && (end = (pos = buf) + fread(buf, 1, 1<<21, stdin), pos == end) ? EOF : *pos ++; }
    inline int rint() {
        register int x = 0, f = 1;register char c;
        while (!isdigit(c = getc())) if (c == '-') f = -1;
        while (x = (x << 1) + (x << 3) + (c ^ 48), isdigit(c = getc()));
        return x * f;
    }
    inline LL rLL() {
        register LL x = 0, f = 1; register char c;
        while (!isdigit(c = getc())) if (c == '-') f = -1;
        while (x = (x << 1ll) + (x << 3ll) + (c ^ 48), isdigit(c = getc()));
        return x * f;
    }
    inline void rstr(char *str) {
        while (isspace(*str = getc()));
        while (!isspace(*++str = getc()))
            if (*str == EOF) break;
        *str = '';
    }
    template<typename T> 
        inline bool chkmin(T &x, T y) { return x > y ? (x = y, 1) : 0; }
    template<typename T>
        inline bool chkmax(T &x, T y) { return x < y ? (x = y, 1) : 0; }    
}
using namespace io;

const int N = 3e5 + 3;
int n;
namespace SA {
    char str[N];
    int sa[N], rk[N], ht[N], M;
    void Rsort(int *buc, int *tp) {
        fill(buc, buc + 1 + M, 0);
        For (i, 1, n) buc[rk[i]] ++;
        For (i, 1, M) buc[i] += buc[i - 1];
        Forr (i, n, 1) sa[ buc[ rk[ tp[i] ] ] -- ] = tp[i];
    }
    void SuffixSort() {
        static int buc[N], tp[N];
        For (i, 1, n) chkmax(M, rk[i] = str[i - 1] - 'a' + 1), tp[i] = i;
        Rsort(buc, tp);
        for (register int len = 1, cnt = 0; cnt < n; len <<= 1, M = cnt) {
            cnt = 0;
            For (i, n - len + 1, n) tp[++cnt] = i;
            For (i, 1, n) if (sa[i] - len > 0) tp[++cnt] = sa[i] - len;
            Rsort(buc, tp); swap(rk, tp); rk[sa[1]] = cnt = 1;
            For (i, 2, n) rk[sa[i]] = (tp[sa[i]] == tp[sa[i - 1]] && tp[sa[i] + len] == tp[sa[i - 1] + len] ? cnt : ++cnt);
        }
    }
    void getHt() {
        int len = 0;
        For (i, 1, n) {
            if (len) len --;
            int j = sa[rk[i] - 1];
            while (str[j + len - 1] == str[i + len - 1]) len ++;
            ht[rk[i]] = len;
        }
    }
}

int size[N], fa[N], id[N];
LL Min[N], Max[N], sum[N], valmax[N], Ans[N];
int find(int x) {
    return x == fa[x] ? x : fa[x] = find(fa[x]);
}
void Merge(int x, int y, int len) {
    x = find(x), y = find(y);
    fa[y] = x;
    sum[len] += 1ll * size[x] * size[y];
    size[x] += size[y];
    chkmax(valmax[x], max(valmax[x], valmax[y]));
    chkmax(valmax[x], max(Min[x] * Min[y], Max[x] * Max[y]));
    chkmin(Min[x], Min[y]);
    chkmax(Max[x], Max[y]);
    chkmax(Ans[len], valmax[x]);
}

int main() {
#ifndef ONLINE_JUDGE
    file("SA");
#endif
    n = rint();
    rstr(SA::str);
    SA::SuffixSort();
    SA::getHt();
    For (i, 1, n) {
        fa[i] = id[i] = i;
        size[i] = 1;
        valmax[i] = Ans[i] = -1e18;
        Min[i] = Max[i] = rint();
    }
    Ans[0] = Ans[n + 1] = -1e18;
    sort(id + 1, id + 1 + n, [] (int a, int b) {return SA::ht[a] > SA::ht[b];});//按照height从大到小排序，id[i]为第i长的height的rk
    For (i, 1, n) 
        Merge(SA::sa[id[i]], SA::sa[id[i] - 1], SA::ht[id[i]]);//合并 height 所关联的两个后缀
    Forr (i, n - 1, 0)
        sum[i] += sum[i + 1], chkmax(Ans[i], Ans[i + 1]);
    For (i, 0, n - 1)
        printf("%lld %lld
", sum[i], !sum[i] ? 0 : Ans[i]);
}