[NOI2015] 品酒大会 题解

[NOI2015] 品酒大会 题解

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题目大意

给定一个长度为 n" role="presentation" style="position: relative;">n$n$ 的字符串 s" role="presentation" style="position: relative;">s$s$ ，和第 i" role="presentation" style="position: relative;">i$i$ 个位置的权值 ai" role="presentation" style="position: relative;">ai$a_i$

对于每一个 r∈[0,n)" role="presentation" style="position: relative;">r∈[0,n)$r\in [0,n)$ ，求满足 lcp(i,j)≥r" role="presentation" style="position: relative;">lcp(i,j)≥r$\text{lcp}(i,j)\ge r$ 的 (i,j)" role="presentation" style="position: relative;">(i,j)$(i,j)$ 的对数

以及所有的 (i,j)" role="presentation" style="position: relative;">(i,j)$(i,j)$ 中， ai×aj" role="presentation" style="position: relative;">ai×aj$a_i\times a_j$ 的最大值

n≤3×105" role="presentation" style="position: relative;">n≤3×105$n\le 3\times {10}^5$

题解

1

可以转化为倒序枚举 r" role="presentation" style="position: relative;">r$r$ 后求 lcp(i,j)=r" role="presentation" style="position: relative;">lcp(i,j)=r$\text{lcp}(i,j)=r$ 的 (i,j)" role="presentation" style="position: relative;">(i,j)$(i,j)$ 数量以及最大值。

求后缀和、 后缀最大值

2

由于 lcp" role="presentation" style="position: relative;">lcp$\text{lcp}$ 是 rk" role="presentation" style="position: relative;">rk$rk$ 上连续的一段 height" role="presentation" style="position: relative;">height$height$ 的最小值

实际上就是要求区间最小值为 r" role="presentation" style="position: relative;">r$r$ 的区间个数

3

将 height" role="presentation" style="position: relative;">height$height$ 降序排序，按照顺序加入。

每次在 p" role="presentation" style="position: relative;">p$p$ 插入一个数，就是合并两堆数，即 p+1" role="presentation" style="position: relative;">p+1$p+1$ 和 p−1" role="presentation" style="position: relative;">p−1$p-1$ 所在堆与 p" role="presentation" style="position: relative;">p$p$ 合并。

用并查集维护大小、最值即可。

code

#include 
using namespace std;
typedef long long LL;
const int N = 3e5 + 5;
int n, sa[N], rk[N], old[N], t[N], id[N], m, h[N];
int fa[N], sz[N];
LL a[N], mx[N], mn[N], s1[N], s2[N], cnt, res = -1e18;
char str[N];
inline void rs() {
	for (int i = 1; i <= m; i++) t[i] = 0;
    for (int i = 1; i <= n; i++) ++t[rk[i]];
    for (int i = 1; i <= m; i++) t[i] += t[i - 1];
    for (int i = n; i >= 1; i--) sa[t[rk[id[i]]]--] = id[i], id[i] = 0;
}
inline int EQ(int x, int y, int k)
{ return old[x] == old[y] && old[x + k] == old[y + k]; }
inline void bui() {
    m = 200;
    for (int i = 1; i <= n; i++) rk[i] = str[i], id[i] = i;
    rs();
    for (int k = 1, p; k <= n; k <<= 1) {
        p = 0;
        for (int i = n - k + 1; i <= n; i++) id[++p] = i;
        for (int i = 1; i <= n; i++) if (sa[i] > k) id[++p] = sa[i] - k;
        rs(), memcpy(old, rk, sizeof(rk)), p = 0;
        for (int i = 1; i <= n; i++) rk[sa[i]] = EQ(sa[i], sa[i - 1], k) ? p : ++p;
        if (p == n) break;
        m = p;
    }
    for (int i = 1, j, k = 0; i <= n; h[rk[i++]] = k)
       	for (k ? --k : 0, j = sa[rk[i] - 1]; str[i + k] == str[j + k]; k++);
}
int fd(int x) {
	while (fa[x] ^ x) x = fa[x] = fa[fa[x]];
	return x;
}
vector<int> b[N];
inline void mer(int x, int y) {
	x = fd(x), y = fd(y);
	cnt += 1ll * sz[x] * sz[y], res = max(res, max(mx[x] * mx[y], mn[x] * mn[y]));
	fa[y] = x, sz[x] += sz[y], mx[x] = max(mx[x], mx[y]), mn[x] = min(mn[x], mn[y]);
}
int main() {
	scanf("%d%s", &n, str + 1);
	for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
	bui();
	for (int i = 1; i <= n; i++) {
		fa[i] = i, sz[i] = 1, mx[i] = mn[i] = a[sa[i]];
		b[h[i]].push_back(i);	
	}
	for (int i = n - 1; ~i; i--) {
//		for (int x : b[i]) mer(x, x - 1);
		int le = b[i].size();
		for (int j = 0; j < le; j++) mer(b[i][j], b[i][j] - 1);
		if (cnt) s1[i] = cnt, s2[i] = res;
	}
	for (int i = 0; i < n; i++) printf("%lld %lld\n", s1[i], s2[i]);
}