hdu 5154 -- Harry and Magical Computer

Harry and Magical Computer

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 641 Accepted Submission(s): 295

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. 1≤n≤100,1≤m≤10000
The next following m lines, each line contains two numbers a b, indicates a dependencies (a, b). 1≤a,b≤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

Sample Output

YES
NO

思路: 用拓扑排序求解.

 1 /*======================================================================
 2  *           Author :   kevin
 3  *            Email :   zerocode.kevin@gmail.com
 4  *         Filename :   HarryAndMagicalComputer.cpp
 5  *       Creat time :   2015-01-23 09:17
 6  *      Description :
 7 ========================================================================*/
 8 #include <iostream>
 9 #include <algorithm>
10 #include <cstdio>
11 #include <cstring>
12 #include <queue>
13 #include <cmath>
14 #define clr(a,b) memset(a,b,sizeof(a))
15 #define INF 0x7f7f7f7f
16 #define M 10005
17 using namespace std;
18 inline int min_32(int (a),int (b)){return (a)<(b)?(a):(b);}
19 inline int max_32(int (a),int (b)){return (a)>(b)?(a):(b);}
20 inline long long min_64(long long (a),long long (b)){return (a)<(b)?(a):(b);}
21 inline long long max_64(long long (a),long long (b)){return (a)>(b)?(a):(b);}
22 int head[M],in[M];
23 struct Node{
24     int to,next;
25 }node[M];
26 void addEdges(int i,int j,int k)
27 {
28     node[k].to = j;
29     node[k].next = head[i];
30     head[i] = k;
31 }
32 int slove(int n)
33 {
34     queue<int>que;
35     for(int i = 1; i <= n; i++){
36         if(in[i] == 0){
37             que.push(i);
38         }
39     }
40     int k,cnt = 0;
41     while(!que.empty()){
42         int t = que.front();
43         que.pop();
44         cnt++;
45         for(k = head[t]; k != -1; k = node[k].next){
46             if(--in[node[k].to] == 0){
47                 que.push(node[k].to);
48             }
49         }
50     }
51     if(cnt == n){
52         return 1;
53     }
54     return 0;
55 }
56 int main(int argc,char *argv[])
57 {
58     int n,m;
59     while(scanf("%d%d",&n,&m)!=EOF){
60         clr(head,-1);
61         clr(in,0);
62         int a,b;
63         for(int i = 0; i < m; i++){
64             scanf("%d%d",&a,&b);
65             addEdges(b,a,i);
66             in[a]++;
67         }
68         if(slove(n)){
69             printf("YES
");
70         }
71         else{
72             printf("NO
");
73         }
74     }
75     return 0;
76 }

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