洛谷P3916 图的遍历

\(\large{题目链接}\)
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\(\Large\textbf{Solution: } \large{1.一种简单的思路：缩点 + dp，这里就不再赘述。\\2.介绍一种O(n + m)的优秀方法，考虑反向建边，然后从n点到1dfs，这样一旦第一次到达一点，那么当前的起点即为这个点的答案。}\)
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\(\Large\textbf{Code: }\)

#include <bits/stdc++.h>
using namespace std;

const int N = 1e5 + 6;

int n, m, vis[N];
vector <int> v[N];

int read() {
	int x = 0;
	char ch = getchar();
	while (!isdigit(ch)) ch = getchar();
	while (isdigit(ch)) x = x * 10 + ch - ‘0‘, ch = getchar();
	return x; 
}

void dfs(int x, int Max) {
	if (vis[x]) return;
	vis[x] = Max;
	for (int i = 0; i < v[x].size(); ++i) dfs(v[x][i], Max);
	return;
}

int main() {
	n = read(), m = read();
	int x, y;
	for (int i = 1; i <= m; ++i) x = read(), y = read(), v[y].push_back(x);
	for (int i = n; i >= 0; --i) dfs(i, i);
	for (int i = 1; i <= n; ++i) printf("%d ", vis[i]); puts("");
	return 0;
} 

洛谷P3916 图的遍历

原文：https://www.cnblogs.com/Miraclys/p/12622750.html