hdu 4324 Triangle LOVE（拓撲排序）

題目連接：http://acm.hdu.edu.cn/showproblem.php?pid=4324

Triangle LOVE

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 3858    Accepted Submission(s): 1516

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

2

5

00100

10000

01001

11101

11000

5

01111

00000

01000

01100

01110

 

Sample Output

Case #1: Yes

Case #2: No

 

Author

BJTU

 

Source

2012 Multi-University Training Contest 3

 

題目大意：給一個n*n的矩陣，Map[i][j]表示i喜歡j；在這個矩陣表示的範圍內，找到是否存在三角戀的關係。

解題思路：作的第一道拓撲排序判環題目。判斷其中是否有環就能夠了，若是存在環必然存在三元環（這個是根據這道題目得出的結論）。

先介紹一下拓撲排序：對一個 有向無環圖(Directed Acyclic Graph簡稱DAG)G進行拓撲排序，是將G中全部頂點排成一個線性序列，使得圖中任意一對頂點u和v，若邊(u,v)∈E(G)，則u在線性序列中出如今v以前。簡單的說，由某個集合上的一個 偏序獲得該集合上的一個 全序，這個操做稱之爲拓撲排序。

再通俗一點就是，一個點一個點的去刪除，刪掉的必須是入度爲0的，若是將入度爲0的所有刪完，還有剩餘邊的話就會產生死鎖，其中必存在環。

 

詳見代碼。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 
 5 using namespace std;
 6 
 7 int indir[2010];//用來表示入度
 8 char Map[2010][2010];//存儲的是i,j兩個節點的關係，1:i love j,0:j love i
 9 
10 int main()
11 {
12     int t;
13     scanf("%d",&t);
14     int ff=1;
15     while (t--)
16     {
17         memset(indir,0,sizeof(indir));
18         int flag=0;
19         int n;
20         scanf("%d",&n);
21         for (int i=0; i<n; i++)
22         {
23             scanf("%s",Map[i]);
24             for (int j=0; j<n; j++)
25             {
26                 if (Map[i][j]=='1')//i love j,也就是i指向j
27                     indir[j]++;//j的入度++
28             }
29         }
30         int j;
31         for (int i=0; i<n; i++)
32         {
33             for (j=0; j<n; j++)
34                 if (indir[j]==0)
35                     break;
36             if (j==n)//若是所有找完都沒有發現入度爲0的必存在環。
37             {
38                 flag=1;
39                 break;
40             }
41             else
42             {
43                 indir[j]--;//j的入度刪掉
44                 for (int k=0;k<n;k++)//j指向的點刪掉
45                 {
46                     if (Map[j][k]=='1')
47                     {
48                         indir[k]--;
49                     }
50                 }
51             }
52         }
53         if (flag==1)
54             printf ("Case #%d: Yes\n",ff++);
55         else
56             printf ("Case #%d: No\n",ff++);
57     }
58 }