There are n SMTP servers connected by network cables. Each of the m cables connects two computers and has a certain latency measured in milliseconds required to send an email message. What is the shortest time required to send a message from server S to server T along a sequence of cables? Assume that there is no delay incurred at any of the servers.

Input
The first line of input gives the number of cases, N. N test cases follow. Each one starts with a line containing n (2<=n<20000), m (0<=m<50000), S (0<=S<n) and T (0<=T<n). S!=T. The next m lines will each contain 3 integers: 2 different servers (in the range [0, n-1]) that are connected by a bidirectional cable and the latency, w, along this cable (0<=w<=10000).

Output
For each test case, output the line "Case #x:" followed by the number of milliseconds required to send a message from S to T. Print "unreachable" if there is no route from S to T.

3
2 1 0 1
0 1 100
3 3 2 0
0 1 100
0 2 200
1 2 50
2 0 0 1

Case #1: 100
Case #2: 150
Case #3: unreachable

最短路问题，拿SPFA练手。

题意：从起点S发送信息到终点T的最短路。直接上SPFA了。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<limits.h>
#include<queue>
using namespace std;
const int INF=0x3f3f3f3f;
const int maxn=20020*5;//RE几次，记得开大点
int head[maxn],end[maxn],cost[maxn];
int next[maxn],d[maxn],cnt[maxn];
int visit[maxn];
int t,n,m,e;
int S,T;
void add(int u,int v,int w)//邻接表
{
    end[e]=v;
    cost[e]=w;
    next[e]=head[u];
    head[u]=e++;
}
void  SPFA(int x)//SPFA
{
    memset(cnt,0,sizeof(cnt));
    memset(visit,0,sizeof(visit));
    memset(d,INF,sizeof(d));
    queue<int>q;
    q.push(x);
    visit[x]=1;
    cnt[x]++;
    d[x]=0;
    q.push(x);
    while(!q.empty())
    {
        int uu=q.front();
//        cout<<q.front()<<endl;
        q.pop();
        visit[uu]=0;
        for(int i=head[uu];i!=-1;i=next[i])
        {
            int vv=end[i];
            int ww=cost[i];
            if(d[vv]>d[uu]+ww)
            {
                d[vv]=d[uu]+ww;
                if(!visit[vv])
                {
                    visit[vv]=1;
                    q.push(vv);
                    if(++cnt[vv]>n)
                        return ;
                }
            }
        }
    }
}
int main()
{
    int u,v,w;
    int cas=0;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d%d%d",&n,&m,&S,&T);
        e=0;
        memset(head,-1,sizeof(head));
        for(int i=1;i<=m;i++)
        {
            scanf("%d%d%d",&u,&v,&w);
            add(u,v,w);//无向图
            add(v,u,w);
        }
        SPFA(S);
//        cout<<cost[0]<<"  "<<cost[1]<<endl;
//       cout<<d[0]<<" "<<d[T]<<endl;
        if(d[T]<INF)  printf("Case #%d: %d\n",++cas,d[T]);
        else  printf("Case #%d: unreachable\n",++cas);
    }
    return 0;
}