HUST 1103 校赛 邻接表-拓扑排序

Posted 2020-07-06 ( m Lweleth)

Description

N students were invited to attend a party, every student has some friends, only if someone’s all friends attend this party, this one can attend the party(ofcourse if he/she has no friends, he/she also can attend it.), now i give the friendship between these students, you need to tell me whether all of them can attend the party.

Note that the friendship is not mature, for instance, if a is b’s friend, but b is not necessary a’s friend. 

 

Input

Input starts with an integer T(1 <= T <= 10), denoting the number of test case.

For each test case, first line contains an integer N(1 <= N <= 100000), denoting the number of students.

Next n lines, each lines first contains an integer K, denoting the number of friends belong to student i(indexed from 1). Then following K integers, denoting the K friends.

You can assume that the number of friendship is no more than 100000, and the relation like student A is himself’s friend will not be existed. 

 

Output

For each test case, if all of the students can attend the party, print Yes, otherwise print No. 

 

Sample Input

2
3
1 2
1 3
1 1
3
1 2
0
1 1

Sample Output

No
Yes

HINT

For the first case:

Student 1 can attend party only if student 2 attend it.

Student 2 can attend party only if student 3 attend it.

Student 3 can attend party only if student 1 attend it.

So no one can attend the party. 

 

题意：给定n个人， 第i个人依赖k个人，只有k个人都参加了，那么i才会参加。问是否所有人都能参加。

思路：数据量有些大，可能无法用并查集判是否有回路解。学长给出方法是用拓扑排序，最后将节点数量和n比较判断即可。

拓扑排序大致是查找出度为零的所有节点，压入队列，一个个弹出，同时将该点和临边删除。

用到了Vector邻接表，发现储存图真是好用。

但是我的代码还有点问题，vector并不需要开二维的，一维就够了。

 1 #include <stdio.h>
 2 #include <iostream>
 3 #include <string.h>
 4 #include <algorithm>
 5 #include <vector>
 6 #include <queue>
 7 #include <utility>
 8 #define  MAXX 100010
 9 using namespace std;
10 const int  INF = 0x3f3f3f3f;
11 pair<vector<int>, int> x;
12 vector< vector<int> >bel(100010);
13 queue< vector<int> >qu;
14 int a[MAXX];
15 int cnt;
16 
17 int findpush(int n)
18 {
19     int flag = 1;
20     for(int i = 1; i <= n; i++)
21     {
22         if(a[i] == 0)
23         {
24             flag = 0;
25             if(bel[i].size())
26                 qu.push(bel[i]);
27             else 
28                 cnt--;
29             a[i] = INF;
30         }
31     }
32     return flag;
33 }
34 
35 int main()
36 {
37     int T, n, t, ed;
38     scanf("%d", &T);
39     while(T--)
40     {
41         scanf("%d",&n);
42         cnt = n;
43 
44         for(int i = 1; i <= n; i++)
45             a[i] = 0, bel[i].clear();
46         for(int i = 1; i <= n; i++)
47         {
48             scanf("%d", &t);
49            /* if(!t)
50                 cnt--;*/
51             while(t--)
52             {
53                 scanf("%d", &ed);
54                 bel[ed].push_back(i);
55                 a[i]++;
56             }
57         }
58         while(qu.empty())
59         {
60             int flag = findpush(n);
61             if(flag == 1)
62             {
63                 if(cnt)
64                     printf("No\n");
65                 else printf("Yes\n");
66                 break;
67             }
68             int temp = qu.size();
69             while(temp--)
70             {
71 
72                 for(int i = 0; i < (qu.front()).size(); i++)
73                 {
74                     a[(qu.front())[i] ]--;
75                 }
76                 qu.pop();
77                 cnt--; 
78             }        
79         }
80     }
81 }