HDU - 1595——find the longest of the shortest（去边最短路）

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.

Input

Each 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.

Output

In 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

题意:

有一城市，这个城市有n个地点和m条连接他们的路，点的编号是从1到n，小X住在1，他想去n。

但是最近正在维修公路，也就是说这m条路有且只有一条是坏的，但是小X不知道是哪一条，一条很关键的路坏了路程就会增加很多，所以小X想知道从1到n *最*坏*情*况* 下的路程。你能帮助他吗?

思路:

用spfa找到1到n的最短路，枚举每条路径，找出去掉每条路径后的最大值。

#include<algorithm>
#include<string.h>
#include<stdio.h>
#include<queue>
#define inf 0x3f3f3f
#define M 1010
using namespace std;
struct node
{
    int v,w;
    int next;
};
node a[2*M*M];
int first[M],path[2*M*M],pa[2*M*M];
int dis[M],vis[M];
int k;
queue<int>q;
void add(int u,int v,int w)//邻接表
{
    a[k].v=v;
    a[k].w=w;
    a[k].next=first[u];
    first[u]=k++;
}
int spfa(int pan)
{
    int x,l;
    memset(dis,0x3f,sizeof(dis));
    memset(vis,0,sizeof(vis));
    dis[1]=0;
    q.push(1);
    while(!q.empty())
    {
        x=q.front();
        vis[x]=0;
        q.pop();
        l=first[x];
        while(l>=0)
        {
            if(dis[a[l].v]>dis[x]+a[l].w)
            {
                dis[a[l].v]=dis[x]+a[l].w;
                if(pan)//记录路径
                {
                    path[a[l].v]=x;
                    pa[a[l].v]=l;
                }
                if(!vis[a[l].v])
                {
                    vis[a[l].v]=1;
                    q.push(a[l].v);
                }
            }
            l=a[l].next;
        }
    }
}
void init()
{
    memset(first,-1,sizeof(first));
    while(!q.empty())
        q.pop();
}
int main()
{
    int n,m,u,v,w,ans;
    while(~scanf("%d%d",&n,&m))
    {
        ans=0;
        init();
        k=1;
        for(int i=0; i<m; i++)
        {
            scanf("%d%d%d",&u,&v,&w);
            add(u,v,w);
            add(v,u,w);
        }
        spfa(1);
        int max_path=inf,sure_path,en=n,l,liu1,liu2;
        while(en!=1)//枚举最短路得每条路径
        {
            l=pa[en];
            l=l+l%2;//因为是双向，处理两条边
            liu1=a[l].w;
            liu2=a[l-1].w;
            a[l].w=inf;
            a[l-1].w=inf;
            spfa(0);
            a[l].w=liu1;
            a[l-1].w=liu2;
            ans=max(ans,dis[n]);
            en=path[en];
        }
        printf("%d\n",ans);
    }
}