UVA 10986 Sending email(SPFA)

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.

Sample InputSample Output

32 1 0 10 1 1003 3 2 00 1 1000 2 2001 2 502 0 0 1

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

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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;}