HDU1595-find the longest of the shortest-dijkstra+记录路径

Marica is very angry with Mirko because he found a new girlfriend and she seeks revenge.Since she doesn't live in the same city, she started preparing for the long journey.We know for every road how many minutes it takes to come from one city to another. 
Mirko overheard in the car that one of the roads is under repairs, and that it is blocked, but didn't konw exactly which road. It is possible to come from Marica's city to Mirko's no matter which road is closed. 
Marica will travel only by non-blocked roads, and she will travel by shortest route. Mirko wants to know how long will it take for her to get to his city in the worst case, so that he could make sure that his girlfriend is out of town for long enough.Write a program that helps Mirko in finding out what is the longest time in minutes it could take for Marica to come by shortest route by non-blocked roads to his city.

InputEach case there are two numbers in the first row, N and M, separated by a single space, the number of towns,and the number of roads between the towns. 1 ≤ N ≤ 1000, 1 ≤ M ≤ N*(N-1)/2. The cities are markedwith numbers from 1 to N, Mirko is located in city 1, and Marica in city N. 
In the next M lines are three numbers A, B and V, separated by commas. 1 ≤ A,B ≤ N, 1 ≤ V ≤ 1000.Those numbers mean that there is a two-way road between cities A and B, and that it is crossable in V minutes.OutputIn the first line of the output file write the maximum time in minutes, it could take Marica to come to Mirko.Sample Input

5 6
1 2 4
1 3 3
2 3 1
2 4 4
2 5 7
4 5 1

6 7
1 2 1
2 3 4
3 4 4
4 6 4
1 5 5
2 5 2
5 6 5

5 7
1 2 8
1 4 10
2 3 9
2 4 10
2 5 1
3 4 7
3 5 10

Sample Output

11
13
27

给定时间是5s，不会超时

题意：

给出的所有路中存在一条路正在修建

要从1走到n，找出最短路径的最大长度

思路：

先一遍dijkstra找到最短路径长度，并且记录路径

再逐个删掉通向最短路径的每一条边换成别的能够到达终点的路

计算其最大长度

难点：在记录路径上

 #include<iostream>
 #include<string.h>
 #include<cmath>
 #include<iomanip>
 #define inf 0x3f3f3f3f
 #define ios() std::ios::sync_with_stdio(false)
 #define cin() cin.tie(0)
 #define cout() cout.tie(0)
 #define mem(a) memset(a,0,sizeof(a))
 using namespace std;

 int vertex[];//记录首先求得的最短路径所经过的顶点
 int n,m;//n顶点总个数，m是边数
 int e[][];
 int dis[],dis1[],dis2[];
 int book[];
 int flag;

 void init()
 {
     for(int i=; i<=n; i++)
     {
         for(int j=; j<=n; j++)
         {
             if(i==j)
                 e[i][j]=;
             else
                 e[i][j]=inf;
             }
     }
 }

 int  dijkstra(int *dis)
 {
     mem(book);mem(dis1);mem(dis2);
     for(int i=;i<=n;i++)
         dis[i]=e[][i];
     book[]=;
     int u;
     for(int i=;i<=n;i++)
     {
         int minn=inf;
         for(int j=;j<=n;j++)
         {
             if(book[j]==&&dis[j]<minn)
             {
                 u=j;
                 minn=dis[j];
             }
         }
         book[u]=;
         for(int k=;k<=n;k++)
         {
             if(e[u][k]<inf&&dis[u]+e[u][k]<dis[k])
             {
                 dis[k]=dis[u]+e[u][k];
                 if(flag)//如果flag=1，则开始记录最短路的路径
                 {
                     vertex[k]=u;//记录顶点
                 }
             }
         }
     }
     return dis[n];
 }

 int main()
 {
     ios();cin();cout();
     int x,y,z;
     while(cin>>n>>m)
     {
         mem(dis);mem(vertex);mem(e);
         init();
         for(int i=; i<m; i++)
         {
             cin>>x>>y>>z;
             e[x][y]=z;
             e[y][x]=z;
         }

         flag=;mem(dis1);
         fill(vertex,vertex+,);
         int shortest=dijkstra(dis1);
       //  cout<<shortest<<"**"<<endl;

 //        for(int i=n;i>=2;i--)
 //            cout<<vertex[i]<<" ";

         flag=;//通过控制flag决定是否记录路径
         int w=n;//一步步倒回去替换掉每一段的距离且变成无穷大，寻找其他边
         int ans=shortest;
         while(w!=)//   通过while去控制是否已经倒退回去遍历到所有的边
         {
           //  cout<<"***"<<endl;
             int frontt=vertex[w];
             int ori=e[frontt][w];
             e[frontt][w]=e[w][frontt]=inf;//第一次进去先变最后一条边
             mem(dis2);
             int ww=dijkstra(dis2);
           //  if(ww!=inf)
                 ans=max(ww,ans);
             e[frontt][w]=e[w][frontt]=ori;
             w=frontt;
            // 变成无穷大之后再进行下一次找的时候可能会需要到这条边，所以需要恢复
         }
         cout<<ans<<endl;
     }
     return ;
 }