poj1125之dijksrta

/*1.对输入做出一定的处理，变成邻接矩阵，是有向图；
2.假设1为源点，求出来1到所有点的最大值，然后以2为源点，求出来到所有的最大值，最后取其最小的值作为
答案；
3.要检查一下是否有不可达的点
这个是maxmin，也就是最大值的最小值
这个题用floyd会很方便但是性能上可能不如dijkstra*/
#include 
#include 
#include 
#include 
#define N 101
#define Inifite 0x3fffffff
#define pli pair//length and index
using namespace std;
int edge[N][N],dist[N],flag[N];
int numOfBrokers;
void initial()
{
	for (int i=1;i<=numOfBrokers;i++)
	{
		for (int j=1;j<=numOfBrokers;j++)
		{
			if(i==j)
				edge[i][j]=0;
			else edge[i][j]=Inifite;
		}
	}
}
void initialDijkstra(int start)
{
	for (int i=1;i<=numOfBrokers;++i)
		dist[i]=Inifite;
	memset(flag,0,sizeof(flag));
	dist[start]=0;
	flag[start]=1;
}
int findIndexOfMax()
{
	int maxIndex=1;
	for (int i=2;i<=numOfBrokers;i++)
	{
		if (dist[i]>dist[maxIndex])
			maxIndex=i;
	}
	return maxIndex;
}
int dijkstra(int start)
{
	initialDijkstra(start);
	priority_queue,greater > q;
	q.push(make_pair(0,start));
	while (!q.empty())
	{
		pli temp=q.top();
		q.pop();
		int minIndex=temp.second;
		int minlength=temp.first;
		flag[minIndex]=1;
		for (int i=1;i<=numOfBrokers;++i)
		{
			if(!flag[i]&&dist[i]>dist[minIndex]+edge[minIndex][i])
			{
				dist[i]=dist[minIndex]+edge[minIndex][i];
				q.push(make_pair(dist[i],i));
			}
		}
	}
	return dist[findIndexOfMax()];
}
int main()
{
	//freopen("in.txt","r",stdin);
	int from,to,costMinutes,j=1;
	while (cin >> numOfBrokers &&numOfBrokers)
	{
		initial();
		//situation just like this:has many times inputs,remember initial some 
		//necessary variations at first,especially counters or outputs
		int minLength=Inifite,minIndex=1,temp,j=1;//every time j will be set as 1;
		while (j<=numOfBrokers)
		{	
			int contracts,i=1;
			cin >> contracts;
			while (i<=contracts)
			{
				from=j;
				cin >> to >> costMinutes;
				edge[from][to]=costMinutes;
				++i;
			}
			++j;
		}
		for (j=1;j<=numOfBrokers;++j)
		{
			temp=dijkstra(j);
			if(temp