求路径中最小值的最大值，应该可以利用网络流的方法解决，不过这道题就利用了dijkstra的方法解决了。
此前POJ 2253利用Floyd方法解决的思路应该也可以应用到这种方法上来

#include <iostream>
#include <algorithm>
#include <queue>
#include <string>
#include <vector>
#include <cstdio>
#include <cstring>
#include <map>
using namespace std;

const int maxn= 1005;
const int INF= 0x3f3f3f3f;

struct Edge
{
	int b, e, d;
};
struct HeapNode
{
	int u, d;
	bool operator < (const HeapNode &rhs) const
	{
		return this->d< rhs.d;
	}
};
int n, m;
int vis[maxn], dis[maxn];
vector<Edge> edges;
vector<int> G[maxn];

void Init()
{
	edges.clear();
	for (int i= 1; i<= n; ++i){
		G[i].clear();
	}
}
void AddEdge(int &b, int &e, int &w)
{
	edges.push_back((Edge){b, e, w});
	int se= edges.size();
	G[b].push_back(se-1);
}
inline void uAddEdge(int &b, int &e, int &w)
{
	AddEdge(b, e, w);
	AddEdge(e, b, w);
}
void Dijkstra(int s)
{
	priority_queue<HeapNode> Q;
	memset(vis, 0, sizeof(vis));
	memset(dis, -1, sizeof(dis));
	dis[s]= INF;
	Q.push((HeapNode){s, 0});

	while (!Q.empty()){
		HeapNode cur= Q.top();
		Q.pop();
		int u= cur.u;
		if (vis[u]){
			continue;
		}
		vis[u]= 1;

		for (vector<int>::iterator g_iter= G[u].begin(); G[u].end()!= g_iter; ++g_iter){
			Edge &eg= edges[*g_iter];
			int d= min(dis[u], eg.d);
			if (d> dis[eg.e]){
				dis[eg.e]= d;
				Q.push((HeapNode){eg.e, d});
			}
		}
	}
}

int main(int argc, char const *argv[])
{
	int kase= 0, sc;
	scanf("%d", &sc);
	while (sc--){
		int b, e, w;
		Init();
		scanf("%d %d", &n, &m);
		for (int i= 0; i< m; ++i){
			scanf("%d %d %d", &b, &e, &w);
			uAddEdge(b, e, w);
		}
		Dijkstra(1);

		printf("Scenario #%d:\n", ++kase);
		printf("%d\n\n", dis[n]);
	}	
	return 0;
}