Uva(10986)

来源：互联网发布：iphone桌面软件管理编辑：程序博客网时间：2024/06/03 17:41

很裸的一道dijk算法题，因为顶点数太多无法用邻接矩阵表示，所以要用临界表来表示

AC代码

#include<stdio.h>#include<string.h>#include<queue>#include<vector> #include<iostream>#include<algorithm>using namespace std;using namespace std;const int maxn=2*50000;#define inf 99999999struct node{    int v;    int u;    int w;    int next;}V[maxn];int head[maxn];int d[maxn];int n,m,s,t;int tol;int done[maxn];typedef pair<int,int>pii; priority_queue<pii,vector<pii>,greater<pii> >q;void init(){    tol=0;    memset(head,-1,sizeof(head));}void dijk(){    memset(done,0,sizeof(done));    for(int i=0;i<=n;i++)    d[i]=inf;    d[s]=0;    q.push(make_pair(d[s],s));    while(!q.empty())    {      pii  u=q.top();      q.pop();      int x=u.second;      if(done[x])      continue;      done[x]=1;      for(int e=head[x];e!=-1;e=V[e].next)       {         int v=V[e].v;         int w=V[e].w;         if(d[x]+w<d[v])         {             d[v]=d[x]+w;             q.push(make_pair(d[v],v));         }      }   }        if(d[t]==inf)      printf("unreachable\n");      else      printf("%d\n",d[t]);}int main(){    int T;    int tmp=0;    scanf("%d",&T);    while(T--)    {    scanf("%d %d %d %d",&n,&m,&s,&t);    init();    m*=2;    int u,v,w;    for(int i=0;i<m;i+=2)    {        scanf("%d %d %d",&u,&v,&w);        V[i].u=u;        V[i].v=v;        V[i].w=w;        V[i].next=head[u];        head[u]=i;        V[i+1].u=v;        V[i+1].v=u;        V[i+1].next=head[v];        V[i+1].w=w;        head[v]=i+1;     }     printf("Case #%d: ",++tmp);    dijk();       }   return 0;    }

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