I

The final battle is coming. Now Harry Potter is located at city 1, and Voldemort is located at city n. To make the world peace as soon as possible, Of course, Harry Potter will choose the shortest road between city 1 and city n. But unfortunately, Voldemort is so powerful that he can choose to destroy any one of the existing roads as he wish, but he can only destroy one. Now given the roads between cities, you are to give the shortest time that Harry Potter can reach city n and begin the battle in the worst case. 

Input

First line, case number t (t<=20). 
Then for each case: an integer n (2<=n<=1000) means the number of city in the magical world, the cities are numbered from 1 to n. Then an integer m means the roads in the magical world, m (0< m <=50000). Following m lines, each line with three integer u, v, w (u != v,1 <=u, v<=n, 1<=w <1000), separated by a single space. It means there is a bidirectional road between u and v with the cost of time w. There may be multiple roads between two cities. 

Output

Each case per line: the shortest time to reach city n in the worst case. If it is impossible to reach city n in the worst case, output “-1”. 

Sample Input

3441 2 52 4 101 3 33 4 8321 2 52 3 10221 2 11 2 2

Sample Output

15-12   

n个点，分别标为1-n，n个点之间有m条路，现在可以毁坏m条路中的任意一条路，让其无法通行，现在让你求最差的情况下的最短路径长度；

分析：刚开始以为先将最短路径的长度求出来，并保存最短路的路径，然后再遍历最短路的每一条边，将其长度改为INF，如此可求得答案，可最后发现这个图是有重边的，只能使用邻接边来存储，并且还要处理好重边 

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <queue>using namespace std;#define N 1010#define M 50050#define INF 1<<28struct Edge{    int v,next,w;///next可以保存连在一起的边在edge中的位置} edge[M*2];///由于是无向边，所以乘以二int n,m,cnt;int head[N];///保存节点N在edge中的idint vis[N],d[N];struct Node{    int pre,id;} used[N];///used[i].pre存储点i的上一个节点,used[i].id存储点i在edge中的地址(都是最短路上的信息）void init(){    cnt=0;    memset(head,-1,sizeof(head));}void addedge(int u,int v,int w){    edge[cnt].v=v;    edge[cnt].w=w;    edge[cnt].next=head[u];    head[u]=cnt++;}void djstl(int s,int t)///第一次是1和0{    for(int i=1; i<=n; i++)    {        d[i]=INF;        if(!t) used[i].pre=s;///s为起点    }    for(int u=head[s]; u!=-1; u=edge[u].next)    {        if(edge[u].w<d[edge[u].v])        {            d[edge[u].v]=edge[u].w;            if(!t) used[edge[u].v].id=u;        }    }    memset(vis,0,sizeof(vis));    vis[s]=1;    for(int i=1; i<=n; i++)    {        int maxn=INF;        int v=-1;        for(int j=1; j<=n; j++)            if(!vis[j]&&maxn>d[j])            {                maxn=d[j];                v=j;            }        if(v==-1) continue;        vis[v]=1;        for(int u=head[v]; u!=-1; u=edge[u].next)        {            int e=edge[u].v;            if(!vis[e]&&d[v]+edge[u].w<d[e])            {                d[e]=d[v]+edge[u].w;                if(!t) used[e].pre=v,used[e].id=u;            }        }    }}int main(){    int s,t,w;    int T;    scanf("%d",&T);    while(T--)    {        scanf("%d %d",&n,&m);        init();        for(int i=0; i<m; i++)        {            scanf("%d %d %d",&s,&t,&w);            addedge(s,t,w);            addedge(t,s,w);        }        djstl(1,0);        int maxn=d[n];        for(int i=n; i!=1; i=used[i].pre)        {            int val=edge[used[i].id].w;///used[i].id在这里的作用非常大，他将提供了最短路在edge中的id，让其能够将其赋值INF            edge[used[i].id].w=INF;            djstl(1,1);            edge[used[i].id].w=val;            maxn=max(maxn,d[n]);        }        if(maxn>=INF) printf("-1\n");        else printf("%d\n",maxn);    }    return 0;}

用邻接表来存储有重边的图，并用了一个used[i].id来记录最短路中的节点i在edge中的id，以便方便讲

该条边的长度赋值为无限大