所有任務合起來構成一個DAG，因為任務可以並行，所以從無前驅的節點到無後繼的節點的所有任務路徑中，最耗時的那一條就是所需要的總時間，因為這樣的搜尋中已經考慮了一個任務所有可能的開始時間。

#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;

int N, c[10005] = {0};
int f[10005];
vector<int> ch[10005];

int dp(int x)
{
	if(f[x]) return f[x];

	const vector<int>& v = ch[x];
	for(int i = 0; i < v.size(); ++i){
		f[x] = max(f[x], dp(v[i]));
	}
	f[x] += c[x];
	return f[x];
}

int main()
{
	int i, k, p;
	while(~scanf("%d", &N)){
	//initialize
		memset(f, 0, (N + 1) << 2);
		for(i = 0; i <= N; ++i) ch[i].clear();
	//input
		for(i = 1; i <= N; ++i){
			scanf("%d", c + i);
			scanf("%d", &k);
			if(k == 0){
				ch[0].push_back(i);
				continue;
			}
			while(k--){
				scanf("%d", &p);
				ch[p].push_back(i);
			}
		}
	//dp
		printf("%d\n", dp(0));
	}
	return 0;
}
 

#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;

int N, c[10005];
int f[10005];
vector<int> ch[10005];
queue<int> q;
bool inq[10005];

int bfs()
{
	memset(inq + 1, false, N);

	int res = 0;
	q.push(0);
	while(!q.empty()){
		int x = q.front(); q.pop(); inq[x] = false;
		const vector<int>& v = ch[x];
		for(int i = 0; i < v.size(); ++i){
			int y = v[i];
			if(f[y] >= f[x] + c[y]) continue;
			f[y] = f[x] + c[y];
			if(ch[y].empty()) res = max(res, f[y]);
			else if(!inq[y]){
				q.push(y);
				inq[y] = true;
			}
		}
	}
	return res;
}

int main()
{
	int i, k, p;
	while(~scanf("%d", &N)){
	//initialize
		memset(f, 0, (N + 1) << 2);
		for(i = 0; i <= N; ++i) ch[i].clear();
	//input
		for(i = 1; i <= N; ++i){
			scanf("%d", c + i);
			scanf("%d", &k);
			if(k == 0){
				ch[0].push_back(i);
				continue;
			}
			while(k--){
				scanf("%d", &p);
				ch[p].push_back(i);
			}
		}
	//bfs
		printf("%d\n", bfs());
	}
	return 0;
}