LOJ 3089: 洛谷 P5319: 「BJOI2019」奥术神杖

题目传送门:LOJ #3089

题意简述:

有一个长度为 \(n\) 的母串,其中某些位置已固定,另一些位置可以任意填。

同时给定 \(m\) 个小串,第 \(i\) 个为 \(S_i\),所有位置都已固定,它的价值为 \(V_i\)

每次每个小串在母串中出现一次,便会给答案的多重集贡献一个 \(V_i\)

最终的答案为多重集的几何平均数,定义空集的几何平均数为 \(1\)

请你求出一个合法母串(往可以填的位置填合法字符)使得答案最大。

\(1\le n,s\le 1501\)\(1\le V_i\le \max V=10^9\),其中 \(\displaystyle s=\sum_{i=1}^{m}|S_i|\)

题解:

假设多重集的大小为 \(c\),第 \(i\) 个元素为 \(w_i\),则 \(\displaystyle\mathrm{Ans}=\sqrt[c]{\prod_{i=1}^{c}w_i}\)

两边取对数,有 \(\displaystyle\ln\mathrm{Ans}=\frac{1}{c}\sum_{i=1}^{c}\ln w_i\),转化为经典的 0/1 分数规划问题。

二分答案,若等式右边大于 \(\mathrm{mid}\),则有:

\(\begin{aligned}\frac{1}{c}\sum_{i=1}^{c}\ln w_i&>\mathrm{mid}\\\sum_{i=1}^{c}\ln w_i&>c\cdot\mathrm{mid}\\\sum_{i=1}^{c}(\ln w_i-\mathrm{mid})&>0\end{aligned}\)

所以,建出小串的 AC 自动机,然后二分答案后在 AC 自动机上 DP 判断不等式是否满足。

DP 时每个小串的权值设为 \(\ln V_i-\mathrm{mid}\),注意要记录最佳转移点,以输出方案。

下面是代码,复杂度 \(\mathcal{O}(s\Sigma(\log\max V-\log\epsilon))\)

#include <cstdio>
#include <cmath>

typedef double f64;
const int MN = 1505, Sig = 10;
const f64 eps = 1e-6, inf = 1e99;

int N, M;
char T[MN];

char str[MN];
int ch[MN][Sig], fail[MN], sum[MN], cnt;
f64 val[MN];

inline void Insert(char *s, f64 v) {
    int now = 0;
    for (; *s; ++s) {
        if (!ch[now][*s & 15]) ch[now][*s & 15] = ++cnt;
        now = ch[now][*s & 15];
    } ++sum[now], val[now] += v;
}

int que[MN], l, r;
void BuildAC() {
    fail[0] = -1;
    que[l = r = 1] = 0;
    while (l <= r) {
        int u = que[l++];
        for (int i = 0; i < Sig; ++i) {
            if (ch[u][i]) {
                int x = fail[u];
                while (~x && !ch[x][i]) x = fail[x];
                if (~x) fail[ch[u][i]] = ch[x][i];
                que[++r] = ch[u][i];
            }
            else if (~fail[u]) ch[u][i] = ch[fail[u]][i];
        }
    }
    for (int i = 2; i <= r; ++i)
        sum[que[i]] += sum[fail[que[i]]],
        val[que[i]] += val[fail[que[i]]];
}

f64 f[MN][MN];
int g[MN][MN][2];
char AT[MN];
inline f64 DP(f64 V) {
    for (int j = 0; j <= cnt; ++j) val[j] -= sum[j] * V;
    for (int i = 0; i <= N; ++i)
        for (int j = 0; j <= cnt; ++j)
            f[i][j] = -inf;
    f[0][0] = 0;
    for (int i = 0; i < N; ++i) {
        for (int j = 0; j <= cnt; ++j) {
            if (f[i][j] == -inf) continue;
            if (T[i] == '.') {
                for (int k = 0; k < Sig; ++k) {
                    int _j = ch[j][k];
                    if (f[i + 1][_j] < f[i][j] + val[_j])
                        f[i + 1][_j] = f[i][j] + val[_j],
                        g[i + 1][_j][0] = j,
                        g[i + 1][_j][1] = k;
                }
            }
            else {
                int _j = ch[j][T[i] & 15];
                if (f[i + 1][_j] < f[i][j] + val[_j])
                    f[i + 1][_j] = f[i][j] + val[_j],
                    g[i + 1][_j][0] = j,
                    g[i + 1][_j][1] = T[i] & 15;
            }
        }
    }
    for (int j = 0; j <= cnt; ++j) val[j] += sum[j] * V;
    int ans = 0;
    for (int j = 1; j <= cnt; ++j)
        if (f[N][j] > f[N][ans]) ans = j;
    for (int i = N, j = ans; i >= 1; --i)
        AT[i - 1] = g[i][j][1] | 48,
        j = g[i][j][0];
    return f[N][ans];
}

int main() {
    scanf("%d%d", &N, &M);
    scanf("%s", T);
    for (int i = 1; i <= M; ++i) {
        f64 v;
        scanf("%s%lf", str, &v);
        Insert(str, log(v));
    }
    BuildAC();
    f64 l = 0, r = log(1e9 + 5), mid, ans = 0;
    while (r - l > eps) {
        mid = (l + r) / 2;
        if (DP(mid) > 0) ans = mid, l = mid;
        else r = mid;
    }
    DP(ans);
    printf("%s\n", AT);
    return 0;
}
posted @ 2019-04-27 21:13 粉兔 阅读(...) 评论(...) 编辑 收藏