LightOJ - 1003 Drunk
阅读原文时间:2023年07月10日阅读:2

One of my friends is always drunk. So, sometimes I get a bit confused whether he is drunk or not. So, one day I was talking to him, about his drinks! He began to describe his way of drinking. So, let me share his ideas a bit. I am expressing in my words.

There are many kinds of drinks, which he used to take. But there are some rules; there are some drinks that have some pre requisites. Suppose if you want to take wine, you should have taken soda, water before it. That's why to get real drunk is not that easy.

Now given the name of some drinks! And the prerequisites of the drinks, you have to say that whether it's possible to get drunk or not. To get drunk, a person should take all the drinks.

Input

Input starts with an integer T (≤ 50), denoting the number of test cases.

Each case starts with an integer m (1 ≤ m ≤ 10000). Each of the next m lines will contain two names each in the format a b, denoting that you must have a before havingb. The names will contain at most 10 characters with no blanks.

Output

For each case, print the case number and 'Yes' or 'No', depending on whether it's possible to get drunk or not.

Sample Input

Output for Sample Input

2

2

soda wine

water wine

3

soda wine

water wine

wine water

Case 1: Yes

Case 2: No


PROBLEM SETTER: JANE ALAM JAN:

思路:直接拓扑排序判断有无环,复杂度O(N+E)

1 #include
2 #include
3 #include
4 #include
5 #include
6 #include
7 #include
8 #include
9 #include
10 #include
11 char str1[20];
12 char str2[20];
13 int cnt[200005];
14 int top(int ans,int nn,int k);
15 using namespace std;
16 vectorvec[20005];
17 int main(void)
18 {
19 int i,j,k,p,q;
20 int s;
21 scanf("%d",&k);
22 for(s=1; s<=k; s++) 23 { 24 mapmy;
25 memset(cnt,0,sizeof(cnt));
26 scanf("%d",&p);
27 int ans=1;
28 for(i=0; i<20004; i++) 29 vec[i].clear(); 30 for(i=0; ique;
55 for(i=1; i<=ans; i++)
56 if(cnt[i]==0)
57 {
58 nn++;
59 que.push(i);
60 }
61 while(!que.empty())
62 {
63 int y=que.front();
64 que.pop();
65 for(i=0; i<vec[y].size(); i++)
66 {
67 cnt[vec[y][i]]--;
68 if(cnt[vec[y][i]]==0)
69 {
70 nn++;
71 que.push(vec[y][i]);
72 }
73 }
74 }
75 if(nn==k)return 1;
76 else return 0;
77 }

手机扫一扫

移动阅读更方便

阿里云服务器
腾讯云服务器
七牛云服务器

你可能感兴趣的文章