1595 hdu find the longest of the shortest

题目地址：

题解：题目意思为从起点到终点的最短路中有一条路不能通过了，求到从起点到终点的最短距离。

可以先找出从起点到终点的最短距离，并将路径保存下来，然后枚举最短路径中的所有路径，求

出从起点到终点的最短路径中最长的一条。

#include<iostream>
#include<string>
#include<queue>
using namespace std;
const int maxn = 1005;
const int INF = 200000000;
struct node
{
	int v,t;
	struct node *next;
}*head[maxn],edge[maxn*maxn];
int n,m,dis[maxn],per[maxn];
bool vis[maxn];
queue<int> que;
void spfa()
{
	for(int j = 1; j <= n; j++)
		dis[j] = INF,vis[j] = false;;
	dis[1] = 0;vis[1] = true;
	que.push(1);
	while(!que.empty())
	{
		int now = que.front();
		que.pop();
		vis[now] = false;
		for(node *p = head[now] ; p ; p = p->next)
		{
			if(dis[p->v] > dis[now] + p->t)
			{
				dis[p->v] = dis[now] + p->t;
				per[p->v] = now;//记录到点p->的前驱结点
				if(!vis[p->v])
				{
					vis[p->v] = true;
					que.push(p->v);
				}
			}
		}
	}
}
void spfa(int x,int y)
{
	for(int i = 1; i <= n; i++)
		dis[i] = INF,vis[i] = false;
	vis[1] = true;
	dis[1] = 0;
	que.push(1);
	while(!que.empty())
	{
		int now = que.front();
		que.pop();
		vis[now] = false;
		for(node *p = head[now] ; p ; p = p->next)
		{
			if((now == y && p->v == x)||(now == x && p->v == y))//路p->v到now的路不能走
				continue;
			if( dis[p->v] > dis[now] + p->t)
			{
				dis[p->v] = dis[now] + p->t;
				if(!vis[p->v])
				{
					vis[p->v] = true;
					que.push(p->v);
				}
			}
		}
	}
}
int main()
{
	int i;
	int a,b,time;
	while(cin >> n >> m)
	{
		for(i = 1; i <= n; i++)
			head[i] = NULL;
		node *p = edge;
		while(m --)
		{
			cin >> a >> b >> time;
			p->v = b;p->t = time;
			p->next = head[a];
			head[a] = p++;
			
			p->v = a; p->t = time;
			p->next = head[b];
			head[b] = p++;
		}
		per[1] = -1;
		spfa();
		int max = dis[n];
		for(i = n; per[i] != -1; i = per[i])//枚举最短路中的所有路径。
		{
			spfa(i,per[i]);
			if(dis[n] != INF && dis[n] > max)
				max = dis[n];
		}
		printf("%d\n",max);
	}
	return 0;
}