[NOI 2015]品酒大会

Description

题库链接

(n) 杯鸡尾酒排成一行，其中第 (i) 杯酒 （(1 leq i leq n)） 被贴上了一个标签 (s_i)，每个标签都是 (26) 个小写英文字母之一。设 (mathrm{Str}(l, r)) 表示第 (l) 杯酒到第 (r) 杯酒的 (r - l + 1) 个标签顺次连接构成的字符串。若 (mathrm{Str}(p, p_o) = mathrm{Str}(q, q_o))，其中 (1 leq p leq p_o leq n)，(1 leq q leq q_o leq n)，(p eq q)，(p_o - p + 1 = q_o - q + 1 = r)，则称第 (p) 杯酒与第 (q) 杯酒是“(r)相似” 的。当然两杯“(r)相似” （(r > 1)）的酒同时也是“(1) 相似”、“(2) 相似”、(dots)、“((r - 1)) 相似”的。特别地，对于任意的 (1 leq p, q leq n)，(p eq q)，第 (p) 杯酒和第 (q) 杯酒都是“(0)相似”的。

如果把第 (p) 杯酒与第 (q) 杯酒调兑在一起，将得到一杯美味度为 (a_p a_q) 的酒。现在对于 (r = 0,1,2, dots, n - 1)，统计出有多少种方法可以选出 (2) 杯“(r)相似”的酒，并回答选择 (2) 杯“(r)相似”的酒调兑可以得到的美味度的最大值。

(1leq nleq 300000)

Solution

可以参考[AHOI 2013]差异这题来统计个数。

找最大值的话，可以用 ( ext{rmq}) 查区间最值相乘。

最后做一遍前缀和以及前缀最大值即可。

Code

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 300000+5;
const ll inf = 1ll*1000000000*1000000000+5;

char ch[N];
int n, m, x[N<<1], y[N<<1], c[N], sa[N], rk[N], height[N], L[N], R[N];
int a[N], val[N], minn[20][N], maxn[20][N], lim, bin[25];
int s[N], top, l[N], r[N], logn[N];
ll cnt[N], ans[N];

void get() {
    for (int i = 1; i <= n; i++) c[x[i] = ch[i]]++;
    for (int i = 2; i <= m; i++) c[i] += c[i-1];
    for (int i = n; i >= 1; i--) sa[c[x[i]]--] = i;
    for (int k = 1; k <= n; k <<= 1) {
        int num = 0;
        for (int i = n-k+1; i <= n; i++) y[++num] = i;
        for (int i = 1; i <= n; i++) if (sa[i] > k) y[++num] = sa[i]-k;
        for (int i = 1; i <= m; i++) c[i] = 0;
        for (int i = 1; i <= n; i++) c[x[i]]++;
        for (int i = 2; i <= m; i++) c[i] += c[i-1];
        for (int i = n; i >= 1; i--) sa[c[x[y[i]]]--] = y[i];
        swap(x, y); x[sa[1]] = num = 1;
        for (int i = 2; i <= n; i++)
            x[sa[i]] = (y[sa[i]] == y[sa[i-1]] && y[sa[i]+k] == y[sa[i-1]+k]) ? num : ++num;
        if ((m = num) == n) break;
    }
    for (int i = 1; i <= n; i++) rk[sa[i]] = i;
    for (int i = 1, k = 0; i <= n; i++) {
        if (rk[i] == 1) continue;
        if (k) --k; int j = sa[rk[i]-1];
        while (i+k <= n && j+k <= n && ch[i+k] == ch[j+k]) ++k;
        height[rk[i]] = k;
    }
}
void rmq() {
    for (int i = 1; i <= n; i++) minn[0][i] = maxn[0][i] = val[i];
    for (int t = 1; t <= lim; t++)
        for (int i = 1; i+bin[t]-1 <= n; i++)
            minn[t][i] = min(minn[t-1][i], minn[t-1][i+bin[t-1]]),
            maxn[t][i] = max(maxn[t-1][i], maxn[t-1][i+bin[t-1]]);
}
int qmin(int l, int r) {
    int t = logn[r-l+1];
    return min(minn[t][l], minn[t][r-bin[t]+1]);
}
int qmax(int l, int r) {
    int t = logn[r-l+1];
    return max(maxn[t][l], maxn[t][r-bin[t]+1]);
}
void work() {
    scanf("%d%s", &n, ch+1); m = 255; logn[0] = -1; bin[0] = 1;
    for (int i = 1; i <= 20; i++) bin[i] = bin[i-1]<<1;
    for (int i = 1; i <= n; i++) logn[i] = logn[i>>1]+1;
    lim = logn[n];
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    get();
    for (int i = 1; i <= n; i++) val[rk[i]] = a[i];
    rmq();
    s[top = 1] = 1;
    for (int i = 2; i <= n; i++) {
        while (top != 1 && height[i] < height[s[top]]) --top;
        l[i] = s[top]; s[++top] = i;
    }
    s[top = 1] = n+1;
    for (int i = n; i >= 2; i--) {
        while (top != 1 && height[i] <= height[s[top]]) --top;
        r[i] = s[top]; s[++top] = i;
    }
    for (int i = 0; i <= n; i++) ans[i] = -inf;
    for (int i = 2; i <= n; i++) {
        cnt[height[i]] += 1ll*(i-l[i])*(r[i]-i);
        ans[height[i]] = max(ans[height[i]], 1ll*qmax(l[i], i-1)*qmax(i, r[i]-1));
        ans[height[i]] = max(ans[height[i]], 1ll*qmin(l[i], i-1)*qmin(i, r[i]-1));
    }
    for (int i = n-1; i >= 0; i--) 
        cnt[i] += cnt[i+1], ans[i] = max(ans[i], ans[i+1]);
    for (int i = n-1; i >= 0; i--)
        if (cnt[i] == 0) ans[i] = 0;
    for (int i = 0; i < n; i++) printf("%lld %lld
", cnt[i], ans[i]);
}
int main() {work(); return 0; }