SOJ 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

1 01 11 13 11 32 21 22 10 88

Sample Output

0 1INVALID1 2INVALID

Problem Source

ZSUACM Team Member

#include<iostream>#include<stdio.h>#include<memory.h>#include<vector>#include<algorithm>using namespace std;  vector<int> a[105];    //存储每个点的邻接表  bool isforest;         // 是否是森林的判断  bool visit[105];       //是否访问过这一节点  int width[105];        //每一层的宽度  int depth,wid;         //森林的深度和宽度  int in[105];           // 保存每个节点的入度，根据定义，入度不能>=2void dfs(int currvet,int level){  if(visit[currvet])  {  isforest=false;  return ;  }   visit[currvet]=true;   width[level]++;       //遍历到了下一层，并且是合法的点，则该层的宽度+1    if(wid<width[level])   wid=width[level];   if(depth<level)   depth=level;               for (int i = 0; i < a[currvet].size(); ++i)   {   /* code */     dfs(a[currvet][i],level+1);   }  }  int main(){int vertex,edge;while(cin>>vertex>>edge,vertex){//初始化操作在下面,在这个循环的i初始值为0       for (int i = 0; i <= vertex; ++i)       {       /* code */          visit[i]=false;          width[i]=0;          a[i].clear();          in[i]=0;       }       wid=depth=0;       isforest=true;       for (int i = 0; i < edge; ++i)       {       /* code */       int first,second;       cin>>first>>second;       a[first].push_back(second);       in[second]++;       if(in[second]>=2)       isforest=false;       }       if(isforest)       {         for (int i = 1; i <= vertex; ++i)         {       /* code */         if(!in[i])                 dfs(i,0);  //如果入度为0，则说明这个是根节点，从第0层开始深搜。         }       }         //针对1 3  这种情况，不是森林，2节点没有被访问到，则为非法的。       for (int i = 1; i <= vertex; ++i)       {       /* code */          if(!visit[i])          isforest=false;       }      if(!isforest)      {    cout<<"INVALID"<<endl;        continue;      }     else       cout<<depth<<" "<<wid<<endl;  }//system("pause");return 0;}