Codeforces 1100E 拓扑排序

题意及思路：https://blog.csdn.net/mitsuha_/article/details/86482347

如果一条边(u, v)，v的拓扑序小于u, 那么(u, v)会形成环，要反向。

代码：

#include <bits/stdc++.h>
#define pii pair<int, int>
using namespace std;
const int maxn = 100010;
vector<int> G[maxn];
vector<int> ans;
int limit;
int deg[maxn], s[maxn];
queue<int> q;
struct edge {
	int x, y, val, id;
};
edge a[maxn];
void add(int x, int y) {
	G[x].push_back(y);
	deg[y]++;
}
int n, m;
bool topsort() {
	ans.clear();
	int tot = 0;
	memset(deg, 0, sizeof(deg));
	for (int i = 1; i <= n; i++)
		G[i].clear();
	for (int i = 1; i <= m; i++) {
		if(a[i].val > limit) {
			add(a[i].x, a[i].y);
		}
	}
	for (int i = 1; i <= n; i++) 
		if(deg[i] == 0) {
			q.push(i);
			s[i] = ++tot;
		}
	while(q.size()) {
		int x = q.front();
		q.pop();
		for (auto y : G[x]) {
			deg[y]--;
			if(deg[y] == 0) {
				q.push(y);
				s[y] = ++tot;
			}
		}
	}
	for (int i = 1; i <= n; i++) if(deg[i] > 0) return 0;
	for (int i = 1; i <= m; i++) {
		if(a[i].val <= limit) {
			if(s[a[i].x] > s[a[i].y])
				 ans.push_back(a[i].id);
		}
	}
	return 1;
}
int main() {
	int x, y, z;
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= m; i++) {
		scanf("%d%d%d", &x, &y, &z);
		a[i] = (edge){x, y, z, i}; 
	}
	int l = 0, r = 1e9;
	while(l < r) {
		limit = (l + r) >> 1;
		bool flag = topsort();
		if(!flag) l = limit + 1;
		else r = limit;
	}
	limit = l;
	topsort();
	printf("%d %d
", l, ans.size());
	for (int i = 0; i < ans.size(); i++)
		printf("%d ", ans[i]);
	printf("
");
}