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60 lines
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2 years ago
#include <bits/stdc++.h>
using namespace std;
const int N = 10010, M = N << 1;
// 链式前向星
int e[M], h[N], idx, w[M], ne[M];
void add(int a, int b, int c = 0) {
e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;
}
int num[N]; // 记录每个工人的最终获取多少奖金
int n, m;
int in[N], ans;
int toposort() {
queue<int> q;
int cnt = 0; // 出队列的节点个数
ans = 0;
for (int i = 1; i <= n; i++)
if (in[i] == 0) q.push(i);
while (q.size()) {
int u = q.front();
q.pop();
cnt++;
ans += num[u]; // 出队时,累加出队人员的奖金数量
for (int i = h[u]; ~i; i = ne[i]) {
int j = e[i];
in[j]--;
if (in[j] == 0) {
q.push(j);
num[j] = num[u] + 1; // 需要表示反向图指定的节点比源节点的奖金大1
}
}
}
if (cnt != n) ans = -1;
return ans;
}
int main() {
// 加快读入
ios::sync_with_stdio(false), cin.tie(0);
while (cin >> n >> m) {
// 初始化链式前向星
memset(h, -1, sizeof h);
idx = 0;
memset(in, 0, sizeof in);
for (int i = 0; i <= n; i++) num[i] = 888; // 每个人最少888元
while (m--) {
int u, v;
cin >> u >> v;
add(v, u); // 小的向大的建图
in[u]++;
}
printf("%d\n", toposort());
}
return 0;
}