计蒜客 28202. Failing Components-最短路(Dijkstra) (BAPC 2014 Preliminary ACM-ICPC Asia Training League 暑假第一阶段第一场 B)

B. Failing Components

传送门

题意就是单向图，从起点开始找最短路，然后统计一下个数就可以。方向是从b到a，权值为s。

直接最短路跑迪杰斯特拉，一开始用数组版的没过，换了一个队列版的过了。

代码:

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<cstdlib>
 #include<queue>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 typedef long long ll;
 const int maxn=1e5+;
 const double eps=1e-;
 const int N=1e5+;
 const int INF=0x3f3f3f3f;
 int head[N*], nex[N*], to[N*], val[N*], dis[N], vis[N], tot;
 struct cmp{
     bool operator()(int a,int b) {
         return dis[a]>dis[b];
     }
 };
 priority_queue<int, vector<int>, cmp > Q;
 void init() {
     tot = ;
     while(!Q.empty()) Q.pop();
     memset(head, -, sizeof(head));
     memset(dis, INF, sizeof(dis));
     memset(vis, , sizeof(vis));
 }
 void addedge(int u, int v, int w) {
     to[tot] = v;
     nex[tot] = head[u];
     val[tot] = w;
     head[u] = tot++;
 }
 void Dijkstra(int S) {
     Q.push(S);
     dis[S] = ;
     while(!Q.empty()) {
         int u = Q.top();
         Q.pop();
         for(int i=head[u]; i!=-; i=nex[i]) {
             int v = to[i];
             if(!vis[v] && dis[u]+val[i] < dis[v]) {
                 dis[v] = dis[u]+val[i];
                 Q.push(v);
             }
         }
     }
 }
 int main(){
     int t;
     scanf("%d",&t);
     while(t--){
         init();
         int n,p,k;
         scanf("%d%d%d",&n,&p,&k);
         int h,l,val;
         for(int i=;i<p;i++){
             scanf("%d%d%d",&h,&l,&val);
             addedge(l,h,val);
         }
         Dijkstra(k);
         int num=;
         int maxx=-;
         for(int i=;i<=n;i++){
             if(dis[i]!=INF){
                 num++;
                 maxx=max(maxx,dis[i]);
             }
         }
         cout<<num<<" "<<maxx;
         cout<<endl;
     }
     return ;
 }