BC Harry and Magical Computer (拓扑排序)

 

 Accepts: 350

 

 Submissions: 1348

 Time Limit: 2000/1000 MS (Java/Others)

 

 Memory Limit: 32768/32768 K (Java/Others)

Problem Description

In reward of being yearly outstanding magic student, Harry gets a magical computer. When the computer begins to deal with a process, it will work until the ending of the processes. One day the computer got n processes to deal with. We number the processes from 1 to n. However there are some dependencies between some processes. When there exists a dependencies (a, b), it means process b must be finished before process a. By knowing all the m dependencies, Harry wants to know if the computer can finish all the n processes.

Input

There are several test cases, you should process to the end of file. For each test case, there are two numbers n m on the first line, indicates the number processes and the number of dependencies. n

Output

Output one line for each test case. If the computer can finish all the process print "YES" (Without quotes). Else print "NO" (Without quotes).

Sample Input

3 2
3 1
2 1
3 3
3 2
2 1
1 3

Sample Output

YES
NO




拓扑排序模板题。注意在处理IN[i] ++的时候要先判断是否已经存在这条边。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <string>
 5 #include <cctype>
 6 #include <cmath>
 7 #include <queue>
 8 #include <map>
 9 #include <cstdlib>
10 using    namespace    std;
11 
12 const    int    SIZE = 105;
13 int    IN[SIZE];
14 bool    G[SIZE][SIZE];
15 int    N,M;
16 
17 bool    toposort(void);
18 int    main(void)
19 {
20     int    from,to;
21 
22     while(scanf("%d%d",&N,&M) != EOF)
23     {
24         fill(IN,IN + SIZE,0);
25         fill(&G[0][0],&G[SIZE - 1][SIZE - 1],false);
26         for(int i = 0;i < M;i ++)
27         {
28             scanf("%d%d",&from,&to);
29             if(!G[from][to])
30                 IN[to] ++;
31             G[from][to] = 1;
32         }
33         if(toposort())
34             puts("YES");
35         else
36             puts("NO");
37     }
38 
39     return    0;
40 }
41 
42 bool    toposort(void)
43 {
44     int    count = 0;
45     queue<int>    que;
46 
47     for(int i = 1;i <= N;i ++)
48         if(!IN[i])
49             que.push(i);
50 
51     while(!que.empty())
52     {
53         int    cur = que.front();
54         que.pop();
55         count ++;
56 
57         for(int i = 1;i <= N;i ++)
58             if(G[cur][i])
59             {
60                 IN[i] --;
61                 if(!IN[i])
62                     que.push(i);
63             }
64     }
65     if(count < N)
66         return    false;
67     return    true;
68 }