from atcoder   

Problem Statement

Time Limit: 2 sec / Memory Limit: 1024 MB

Score : 300300 points

Takahashi is a martial artist. There are NN moves that a martial artist can learn, called Move 11, 22, \ldots…, NN. For each 1 \leq i \leq N1≤i≤N, it takes T_iTi​ minutes of practice to learn Move ii. Additionally, at the beginning of that practice, all of the Moves A_{i,1}Ai,1​, A_{i,2}Ai,2​, \ldots…, A_{i,K_i}Ai,Ki​​ must be already learned. Here, it is guaranteed that A_{i,j} < iAi,j​<i for each 1 \leq j \leq K_i1≤j≤Ki​.

Takahashi has not learned any move at time 00. He cannot practice for more than one move simultaneously, nor can he stop a practice he has already started. Find the minimum number of minutes needed for Takahashi to learn Move NN.

Constraints

• 

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Input

Input is given from Standard Input in the following format:

Output

Print the minimum number of minutes needed for Takahashi to learn Move N.

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Sample Input 1 Copy

Copy

3
3 0
5 1 1
7 1 1

Sample Output 1 Copy

Copy

10

Here is one possible plan for Takahashi.

• At time 00, start practicing for Move 11 to learn Move 11 at time 33.
• Then, at time 33, start practicing for Move 33 to learn Move 33 at time 1010.

Here, Takahashi spends 3+7=103+7=10 minutes to learn Move 33, which is the fastest possible. Note that he does not need to learn Move 22 to learn Move 33.

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Sample Input 2 Copy

Copy

5
1000000000 0
1000000000 0
1000000000 0
1000000000 0
1000000000 4 1 2 3 4

Sample Output 2 Copy

Copy

5000000000

Note that the answer may not fit into a 32-bit integer.

解题目标：获得学会动作Move N所需的最短时间；

题目大义：一个人要学动作，在学一个动作之前要学会一些当前动作要求的基础动作，只要学会基础动作中的一个，就可以学习当前动作了，本题input给出N个动作数量，以及学习每个动作的耗时与每个动作的基础动作，求最快学到动作N的时间；

题目类型：图，搜索，bfs,求深度和值之和。

注意点：基础动作可能存在重复情况，已经学过的动作不用重复学。

解题思路：

        1）用一个邻接表来记录当前第i个Move所需的基础动作；

        2）用t[]记录学第i个动作的耗时；

        3）用vi[]来记录当前动作是否以及学过；

        4)利用queue 边判断边加时间边压；

#include <bits/stdc++.h>
//#define MAXN 2e5+50
#define rep(x, a, b) for(int x=a; x<=b; x++)
#define per(x, a, b) for(int x=a; x>=b; x--)
using namespace std;
const int MAXN = 2e5+10;
vector<int> group[MAXN];
bool vi[MAXN];
int t[MAXN];

int main()
{
    int n;
    scanf("%d", &n);
    rep(i, 1, n)
    {
        int len;
        scanf("%d%d",&t[i] , &len);
      //  cout<<t[i]<< " "<<len<<endl;
        rep(j,0, len-1)
        {
            int x;
            scanf("%d", &x);
            group[i].push_back(x);
        }
    }
    vi[n] = true;
    queue<int> que;
    que.push(n);
    long long ans = 0 ;
    while(!que.empty())
    {
        int c = que.front();
        que.pop();
        ans += t[c];
        rep(i, 0, (int)(group[c].size() - 1))
        {
            if(vi[group[c][i]]) continue;
            else{
                vi[group[c][i]] = 1;
                que.push(group[c][i]);
            }
        }
    }

    cout<<ans;
	return 0;
}