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
 Copy sample input to clipboard
1 0
1 1
1 1
3 1
1 3
2 2
1 2
2 1
0 88
Sample Output
0 1
INVALID
1 2
INVALID
  
   
View Code 
 1 #include <iostream>
 2 #include <vector>
 3 #include <cstring>
 4 using namespace std;
 5 
 6 #define Max 102
 7 vector<int>v[Max];
 8 int allWidth[Max];
 9 bool visited[Max];
10 bool end[Max];
11 bool forest;
12 int n, m;
13 int depth, width;
14 void dfs(int a, int d)
15 {
16     if (!forest)
17         return;
18     visited[a] = true;
19     if (depth < d)
20         depth = d;
21     if (width < ++allWidth[d])
22         width = allWidth[d];
23     int size = v[a].size();
24     for (int i = 0; i < size; i++){
25         if (visited[v[a][i]]){
26             forest = false;
27             return;                      
28         }else{
29             visited[v[a][i]] = true;
30             dfs(v[a][i], d+1);     
31         }
32     }    
33 }
34 
35 int main()
36 {
37     int a, b;
38     while (cin >> n >> m && n){
39         for (int i = 1; i <= n; i++)
40             v[i].clear();
41         memset(visited, false, sizeof(visited));
42         memset(end, false, sizeof(end));  
43         memset(allWidth, 0, sizeof(allWidth));
44         depth = 0;
45         width = 0;
46         forest = true;
47           
48         for (int i = 0; i < m; i++){
49             cin >> a >> b;
50             v[a].push_back(b);
51             if (end[b] || a == b)
52                 forest = false;
53             else
54                 end[b] = true;
55         }
56         if (!forest){
57             cout << "INVALID" << endl;
58             continue;
59         }
60         for (int i = 1; i <= n; i++){
61             if (!end[i]){
62                dfs(i, 0);             
63             }    
64         }
65         for (int i = 1; i <= n; i++){
66             if (!visited[i]){
67                forest = false;
68                break;                 
69             }
70         } 
71         if (forest)
72             cout << depth << " " << width << endl;
73         else
74             cout << "INVALID" << endl;
75     }   
76     
77     //system("PAUSE");
78     return 0;   
79 }

 

posted @ 2012-12-22 13:39  gumcstronger  阅读(370)  评论(0)    收藏  举报