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;
}