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+10;
const double eps=1e-7;
const int N=1e5+10;
const int INF=0x3f3f3f3f;
int head[N*2], nex[N*2], to[N*2], val[N*2], 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 = 0;
    while(!Q.empty()) Q.pop();
    memset(head, -1, sizeof(head));
    memset(dis, INF, sizeof(dis));
    memset(vis, 0, 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] = 0;
    while(!Q.empty()) {
        int u = Q.top();
        Q.pop();
        for(int i=head[u]; i!=-1; 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=0;i<p;i++){
            scanf("%d%d%d",&h,&l,&val);
            addedge(l,h,val);
        }
        Dijkstra(k);
        int num=0;
        int maxx=-1;
        for(int i=1;i<=n;i++){
            if(dis[i]!=INF){
                num++;
                maxx=max(maxx,dis[i]);
            }
        }
        cout<<num<<" "<<maxx;
        cout<<endl;
    }
    return 0;
}