sicily 1034 Forest

Description

In the field of computer science, forest is important and deeply researched , it is a model for many data structures . Now it’s your job here to calculate the depth and width of given forests.

Precisely, a forest here is a directed graph with neither loop nor two edges pointing to the same node. Nodes with no edge pointing to are roots, we define that roots are at level 0 . If there’s an edge points from node A to node B , then node B is called a child of node A , and we define that B is at level (k+1) if and only if A is at level k .

We define the depth of a forest is the maximum level number of all the nodes , the width of a forest is the maximum number of nodes at the same level.

Input

There’re several test cases. For each case, in the first line there are two integer numbers n and m (1≤n≤100, 0≤m≤100, m≤n*n) indicating the number of nodes and edges respectively , then m lines followed , for each line of these m lines there are two integer numbers a and b (1≤a,b≤n)indicating there’s an edge pointing from a to b. Nodes are represented by numbers between 1 and n .n=0 indicates end of input.

Output

For each case output one line of answer , if it’s not a forest , i.e. there’s at least one loop or two edges pointing to the same node, output “INVALID”(without quotation mark), otherwise output the depth and width of the forest, separated by a white space.

Sample Input

Sample Output

0 1
INVALID
1 2
INVALID

分析：

本题要求森林的深度和宽度，建议用深搜，因为广搜的循环控制条件很难写，极易出错或者死循环。要注意图生成的森林中，可能会出现公共子节点以及环等不合法情况，要注意判断条件。DFS过程中遇到了已经访问过的节点说明有公共子节点的存在，DFS完成后仍有节点未被访问说明有环存在。另外，一次完整的DFS过程中求得的广度只是一棵树的广度，要对同一深度的节点全部统计才能得到森林的广度。

代码：

 1 // Problem#: 1034
 2 // Submission#: 1795986
 3 // The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License
 4 // URI: http://creativecommons.org/licenses/by-nc-sa/3.0/
 5 // All Copyright reserved by Informatic Lab of Sun Yat-sen University
 6 #include <iostream>
 7 #include <cstring>
 8 using namespace std;
 9 
10 #define N 101
11 #define M 101
12 
13 bool flag;
14 bool buffer[N][N];
15 bool visit[N];
16 bool isroot[N];
17 int maxd,n,m,maxb;
18 int b[N];
19 
20 void dfs( int p, int td ){
21     if( visit[p] ){
22         flag = false;
23         return ;
24     }
25     visit[p] = true;
26     maxd = td>maxd ? td : maxd;
27     if(++b[td]>maxb) maxb = b[td];
28     for( int i=1 ; i<=n ; i++ ){
29         if( buffer[p][i] ){
30             if(!flag) return ;
31             dfs(i,td+1);
32         }else
33             continue;
34     }
35 }
36 
37 int main(){
38     int x,y;
39     while( cin>>n>>m && n ){
40         memset(buffer,0,sizeof(buffer));
41         memset(visit,0,sizeof(visit));
42         memset(isroot,true,sizeof(isroot));
43         flag = true;
44         if(m>=n) flag = false;
45         for( int i=0 ; i<m ; i++ ){
46             cin >> x >> y;
47             buffer[x][y] = true;
48             isroot[y] = false;
49         }
50         maxd = maxb = 0;
51         memset(b,0,sizeof(b));
52         for( int i=1 ; i<=n ; i++ )
53             if( isroot[i] ) dfs(i,0);
54         for( int i=1 ; i<=n ; i++ ){
55             if( !visit[i] ){
56                 flag = false;
57                 break;
58             }
59         }
60         if( flag ) cout << maxd << " " << maxb;
61         else cout << "INVALID";
62         cout << endl;
63     }
64     return 0;
65 }