[NOI 2015]品酒大会

时间：2023-03-08 17:29:45
[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 \(1 \leq p \leq p_o \leq n\)，\(1 \leq q \leq q_o \leq n\)，\(p \neq q\)，\(p_o - p + 1 = q_o - q + 1 = r\)，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 \(1 \leq p, q \leq n\)，\(p \neq q\)，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。

\(1\leq n\leq 300000\)

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 \(\text{rmq}\) 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\n", cnt[i], ans[i]);

}

int main() {work(); return 0; }

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