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题目

2773. 函数调用

算法标签: 拓扑排序, 组合计数

思路

算每个函数最终对答案的贡献是多少, 计算当期加法但是要计算后面乘法的贡献, 用乘法原理和加法原理计算每个函数等效的执行次数, 时间复杂度是线性的$O(n)$

为什么$mul$需要按照拓扑序的逆序计算?
假设函数$A$调用了函数$B$和函数$C$, 那么$A$的最终的$mul$是取决于$B$和$C$的$mul$的, 因此需要按照拓扑序的逆序倒着推, 类似于动态规划的过程, 确保当前状态已经被计算

代码

#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>using namespace std;typedef long long LL;
const int N = 1e5 + 10, M = 1e6 + 10, MOD = 998244353;int n, m, q_num;
vector<int> head[N];
int w[N], g[N];
int q[N], deg[N];
struct Func {int t, pos, val;int mul;int sum;
} f[N];void add(int u, int v) {head[u].push_back(v);
}void top_sort() {int h = 0, t = -1;for (int i = 1; i <= m; ++i) {if (deg[i] == 0) q[++t] = i;}while (h <= t) {int u = q[h++];for (int v: head[u]) {if (--deg[v] == 0) q[++t] = v;}}
}void calc_mul() {for (int i = m - 1; i >= 0; --i) {int u = q[i];for (int v : head[u]) {f[u].mul = (LL) f[u].mul * f[v].mul % MOD;}}
}void calc_sum() {for (int i = 0; i < m; ++i) {int u = q[i];int sum = f[u].sum;for (int v : head[u]) {f[v].sum = (f[v].sum + sum) % MOD;sum = (LL) sum * f[v].mul % MOD;}}
}int main() {ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);cin >> n;for (int i = 1; i <= n; ++i) cin >> w[i];cin >> m;for (int i = 1; i <= m; ++i) {cin >> f[i].t;if (f[i].t == 1) cin >> f[i].pos >> f[i].val;else if (f[i].t == 2) cin >> f[i].val;else {int cnt;cin >> cnt;vector<int> tmp;while (cnt--) {int x;cin >> x;tmp.push_back(x);}// 为了保证结果是正确的, 需要从最后一个子函数向前计算reverse(tmp.begin(), tmp.end());for (int x: tmp) {add(i, x);deg[x]++;}}}// 初始化乘法因子for (int i = 1; i <= m; ++i) {if (f[i].t == 2) {f[i].mul = f[i].val;}else {f[i].mul = 1;}}top_sort();calc_mul();cin >> q_num;for (int i = 1; i <= q_num; ++i) cin >> g[i];// 计算全局的乘法标记和加法标记int sum = 1;for (int i = q_num; i >= 1; --i) {int k = g[i];f[k].sum = (f[k].sum + sum) % MOD;sum = (LL) sum * f[k].mul % MOD;}calc_sum();// 应用全局乘法for (int i = 1; i <= n; ++i) {w[i] = (LL) w[i] * sum % MOD;}// 处理类型1的加法操作for (int i = 1; i <= m; ++i) {if (f[i].t == 1) {w[f[i].pos] = (w[f[i].pos] + (LL) f[i].val * f[i].sum) % MOD;}}for (int i = 1; i <= n; ++i) cout << w[i] << " ";cout << "\n";return 0;
}