原题链接:忙碌的小L
hqz nb
#include <bits/stdc++.h> constexpr auto Inf = 0X3F3F3F3F; using namespace std; typedef long long LL; namespace IO { inline LL read() { LL o = 0, f = 1; char c = getchar(); while (c < '0' || c > '9') { if (c == '-') f = -1; c = getchar(); } while (c > '/' && c < ':') { o = o * 10 + c - '0'; c = getchar(); } return o * f; } inline char recd() { char o; while ((o = getchar()) != 'G' && o != 'C'); return o; } } using namespace IO; const int SIZE = 1E5 + 7; vector<int> G[SIZE]; int dfn[SIZE], scc[SIZE], low[SIZE], dfn_index, scc_index; int on_stack[SIZE], top, stak[SIZE]; void tarjan(int u) { on_stack[u] = 1, stak[++top] = u; dfn[u] = low[u] = ++dfn_index; for (int e = 0; e < G[u].size(); e++) { int v = G[u][e]; if (!dfn[v]) tarjan(v), low[u] = min(low[u], low[v]); else if (on_stack[v]) low[u] = min(low[u], dfn[v]); } if (low[u] == dfn[u]) { ++scc_index; while (stak[top] != u) on_stack[stak[top]] = 0, scc[stak[top--]] = scc_index; on_stack[stak[top]] = 0, scc[stak[top--]] = scc_index; } } int indegree[SIZE]; int main() { int N = read(); for (int u = 1, v; u <= N; u++) while (233) { v = read(); if (!v) break; G[u].push_back(v); } for (int u = 1; u <= N; u++) if (!dfn[u]) tarjan(u); for (int u = 1; u <= N; u++) for (int e = 0; e < G[u].size(); e++) { int v = G[u][e]; if (scc[u] != scc[v]) indegree[scc[v]]++; } int ans = 0; for (int p = 1; p <= scc_index; p++) ans += indegree[p] == 0; printf("%d\n", ans); }
0.0分
0 人评分
C语言网提供由在职研发工程师或ACM蓝桥杯竞赛优秀选手录制的视频教程,并配有习题和答疑,点击了解:
一点编程也不会写的:零基础C语言学练课程
解决困扰你多年的C语言疑难杂症特性的C语言进阶课程
从零到写出一个爬虫的Python编程课程
只会语法写不出代码?手把手带你写100个编程真题的编程百练课程
信息学奥赛或C++选手的 必学C++课程
蓝桥杯ACM、信息学奥赛的必学课程:算法竞赛课入门课程
手把手讲解近五年真题的蓝桥杯辅导课程
发表评论 取消回复