[sicily online]1034. Forest

Constraints

Time Limit: 1 secs, Memory Limit: 32 MB

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 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

题目分析：

大意就是判断一个森林是否成立，并且求出该森林的宽度（横向）和深度（纵向）

我是用map来存储图，key是标示，value是与它相连的点的集合，通过广度优先遍历（队列）来求出每个节点所在的层，然后就很容易求出宽度和深度了

#include<iostream>
#include<stdio.h>
#include<cmath>
#include<iomanip>
#include<list>
#include <map>
#include <vector>
#include <string>
#include <algorithm>
#include <sstream>
#include <stack>
#include<queue>
#include<string.h>
using namespace std;

typedef struct INFO
{
	bool isUsed;
	bool isRoot;
	int layer;
	vector<int> child;
}node;

int main()
{
	int n,m;
	while(cin>>n>>m&&n!=0)
	{
		map<int, node>gra;//map应该不能指定尺寸，因为默认值都一样，map是不准key一样的
		for(int i=0;i<n;i++)
		{//处理点
			node tmp;
			tmp.isUsed=false;
			tmp.isRoot=true;
			tmp.layer=0;
			gra[i]=tmp;
		}
		for(int i=0;i<m;i++)
		{//处理边
			int x1,x2;
			cin>>x1>>x2;
			gra[--x1].child.push_back(--x2);
			gra[x2].isRoot=false;
		}

		vector<int> root;//统计根节点
		for(int i=0;i<n;i++)
			if(gra[i].isRoot)
				root.push_back(i);
		int count=0;
		int flag=false;
		for(int i=0;i<root.size();i++)
		{
			queue<int> st;
			count++;
			if(gra[root[i]].isUsed)
			{
				//cout<<"INVALID"<<endl;
				flag=true;
				break;
			}
			gra[root[i]].isUsed=true;
			st.push(root[i]);
			while(!st.empty())
			{
				int index=st.front();
				st.pop();
				vector<int>::iterator ite;
				for(ite=gra[index].child.begin();ite!=gra[index].child.end();ite++)
				{
					if(gra[*ite].isUsed)
					{
						//cout<<"INVALID"<<endl;
						flag=true;
						break;
					}
					else
					{
						count++;
						gra[*ite].isUsed=true;
						gra[*ite].layer=gra[index].layer+1;
						st.push(*ite);
					}//end else
				}//edn for
				if(flag)break;
			}//end while
		}//end for i
		int width=1,height=0;
		map<int,int> tmpWidth;
		for(int i=0;i<n;i++)
		{
			tmpWidth[gra[i].layer]++;
			if(gra[i].layer>height)
				height=gra[i].layer;
		}
		for(map<int,int>::iterator ite=tmpWidth.begin();ite!=tmpWidth.end();ite++)
			if(width<ite->second)width=ite->second;
		if(count<n||flag)
			cout<<"INVALID"<<endl;
		else cout<<height<<" "<<width<<endl;
	}//end whie
}