杭电4324--Triangle LOVE（拓扑排序）

Triangle LOVE

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3479 Accepted Submission(s): 1350

Problem Description

Recently, scientists find that there is love between any of two people. For example, between A and B, if A don’t love B, then B must love A, vice versa. And there is no possibility that two people love each other, what a crazy world!
Now, scientists want to know whether or not there is a “Triangle Love” among N people. “Triangle Love” means that among any three people (A,B and C) , A loves B, B loves C and C loves A.
Your problem is writing a program to read the relationship among N people firstly, and return whether or not there is a “Triangle Love”.

Input

The first line contains a single integer t (1 <= t <= 15), the number of test cases.
For each case, the first line contains one integer N (0 < N <= 2000).
In the next N lines contain the adjacency matrix A of the relationship (without spaces). A_i,j = 1 means i-th people loves j-th people, otherwise A_i,j = 0.
It is guaranteed that the given relationship is a tournament, that is, A_i,i= 0, A_i,j ≠ A_j,i(1<=i, j<=n,i≠j).

Output

For each case, output the case number as shown and then print “Yes”, if there is a “Triangle Love” among these N people, otherwise print “No”.
Take the sample output for more details.

Sample Input

Sample Output

Case #1: Yes 

Case #2: No

Author

BJTU

Source

2012 Multi-University Training Contest 3

Recommend

zhoujiaqi2010 | We have carefully selected several similar problems for you: 1068 4320 4325 3309 3712

RE:三角恋 → → 判断是否成环。

 1 #include <queue>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <iostream>
 5 using namespace std;
 6 char map[2020][2020]; 
 7 int indegree[2020];
 8 int n; 
 9 bool Solve()
10 {
11     queue<int> q;
12     for(int i = 0; i < n; i++)
13     {
14         if(indegree[i] == 1)
15             q.push(i);
16     }
17     int num = 0;
18     while(!q.empty())
19     {
20         int temp = q.front();
21         q.pop();
22         num++;
23         for(int i = 0; i < n; i++)
24         {
25             if(map[temp][i])
26             {
27                 indegree[i]--;
28                 if(indegree[i] == 0)    
29                 q.push(i);
30             }
31         } 
32     }
33     if(num == n)
34         return false;
35     else
36         return true;
37 } 
38 int main()
39 {
40     int t, temp = 1;
41     scanf("%d", &t);
42     while(t--)
43     {
44         scanf("%d", &n);
45         memset(indegree, 0, sizeof(indegree));
46         for(int i = 0; i < n; i++)
47         {
48             scanf("%s", map[i]);
49             for(int j = 0; j < n; j++)
50                 if(map[i][j] == '1')
51                     indegree[j]++;
52         }
53         if(Solve())
54             printf("Case #%d: Yes
", temp++);
55         else
56             printf("Case #%d: No
", temp++);
57     } 
58     return 0;    
59 }

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <iostream>
 4 using namespace std;
 5 char map[2020][2020], indegree[2020];
 6 int n;
 7 bool Tsort()
 8 {
 9     int m;
10     for(int i = 0; i < n; i++)
11     {
12         m = -1;
13         for(int j = 0; j < n; j++)
14         {
15             if(indegree[j] == 0)
16             {
17                 m = j;
18                 break;
19             }
20         }
21         if(m == -1)
22             return true;
23         indegree[m]--;
24         for(int j = 0; j < n; j++)
25         {
26             if(map[m][j] == '1')
27                 indegree[j]--;
28         }
29     }
30     return false;
31 }
32 int main()
33 {
34     int t, Q = 1;
35     scanf("%d", &t);
36     while(t--)
37     {
38         memset(indegree, 0, sizeof(indegree));
39         scanf("%d", &n);
40         for(int i = 0; i < n; i++)
41         {
42             scanf("%s", map[i]);
43             for(int j = 0; j < n; j++)
44                 if(map[i][j] == '1')
45                     indegree[j]++;    
46         }    
47         if(Tsort())
48             printf("Case #%d: Yes
", Q++);
49         else
50             printf("Case #%d: No
", Q++);
51     }
52     return 0;
53 }

数组模拟