二分,每次取到一个 \(mid\) ,只保留长度 \(>mid\) 的边
dfs判环,若有环,说明 \(ans>mid\) ,否则 \(ans≤mid\)
找到 \(ans\) 后,对长度 \(>ans\) 的边进行一次拓扑排序,对多余的点直接接在拓扑排序序列后面即可
对从 \(x\) 到 \(y\) 的长度 \(≤ans\) 的边,如果 \(x\) 在拓扑排序序列中的位置比 \(y\) 后,则这条边需取反
时间复杂度 \(O(n\ log\ C)\)
#include <bits/stdc++.h>
using namespace std;
const int N = 100006;
int n, m, mx = 0, d[N], b[N], t;
int Head[N], Edge[N], Leng[N], Next[N], Pose[N];
vector<int> ans;
bool v[N], w[N];
queue<int> q;
inline void add(int x, int y, int z, int i) {
Edge[i] = y;
Leng[i] = z;
Next[i] = Head[x];
Head[x] = i;
Pose[i] = x;
}
bool dfs(int x, int now) {
v[x] = 1;
w[x] = 1;
for (int i = Head[x]; i; i = Next[i]) {
int y = Edge[i], z = Leng[i];
if (z <= now) continue;
if (w[y] || !dfs(y, now)) return 0;
}
w[x] = 0;
return 1;
}
inline bool pd(int now) {
memset(v, 0, sizeof(v));
memset(w, 0, sizeof(w));
for (int i = 1; i <= n; i++)
if (!v[i] && !dfs(i, now)) return 0;
return 1;
}
void topsort(int now) {
for (int i = 1; i <= n; i++)
if (!d[i]) q.push(i);
while (q.size()) {
int x = q.front();
q.pop();
b[x] = ++t;
for (int i = Head[x]; i; i = Next[i]) {
int y = Edge[i], z = Leng[i];
if (z > now && !--d[y]) q.push(y);
}
}
}
unsigned int work(int now) {
for (int i = 1; i <= m; i++) {
int y = Edge[i], z = Leng[i];
if (z > now) ++d[y];
}
topsort(now);
for (int i = 1; i <= n; i++)
if (!b[i]) b[i] = ++t;
for (int i = 1; i <= m; i++) {
int x = Pose[i], y = Edge[i], z = Leng[i];
if (z <= now && b[x] > b[y]) ans.push_back(i);
}
return ans.size();
}
int main() {
cin >> n >> m;
for (int i = 1; i <= m; i++) {
int x, y, z;
scanf("%d %d %d", &x, &y, &z);
add(x, y, z, i);
mx = max(mx, z);
}
int l = 0, r = mx;
while (l < r) {
int mid = (l + r) >> 1;
if (pd(mid)) r = mid;
else l = mid + 1;
}
cout << l << " " << work(l) << endl;
for (unsigned int i = 0; i < ans.size(); i++)
printf("%d ", ans[i]);
return 0;
}