The 2019 Asia Nanchang First Round Online Programming Contest Fire-Fighting Hero (Dijkstra)

 

Fire-Fighting Hero

 

It has been built not think of a point, into a single source shortest.

Then, it becomes a problem Dijkstra water.

 

#include<bits/stdc++.h> 
using namespace std;
typedef long long ll;
const int mod=998244353;
const int maxn=1e3+50;
const int inf=0x3f3f3f3f;

struct P{
    int to,v;
    P(){}
    P(int tto,int vv){to=tto;v=vv;}
};
vector<P> p[maxn];

struct up{
    bool operator()(const P&p1,const P&p2)const{return p1.v>p2.v;}
};//单调递增 
//star to (que[].to) cost (que[].v);

int vpoi[maxn];
ll dis[maxn],n,maxval;
void Dij(int s)
{
    priority_queue<P,vector<P>,up> que;
    memset(dis,inf,sizeof(dis));
    int i,j;
    int vis[maxn],pre[maxn],k;
    P poi;
    for(i=1;i<=n;i++)pre[i]=s,vis[i]=0;
    for(i=0;i<p[s].size();i++)
    {
        dis[p[s][i].to]=p[s][i].v;
        que.push(P(p[s][i].to,dis[p[s][i].to]));
    }
    dis[s]=0;vis[s]=1;pre[s]=0;
    for(i=1;i<n;i++)
    {
        poi=que.top();que.pop();
        while(dis[poi.to]!=poi.v||vis[poi.to])//失效或已访问 
        {
            poi=que.top();que.pop();
        }
        k=poi.to;vis[k]=1;
        for(j=0;j<p[k].size();j++)
        {
            if(!vis[p[k][j].to]&&dis[p[k][j].to]>dis[k]+p[k][j].v)
            {
                dis[p[k][j].to]=dis[k]+p[k][j].v;
                pre[p[k][j].to]=k;
                que.push(P(p[k][j].to,dis[p[k][j].to]));//更新 
            }
        }
    }
    maxval=0;
    for(i=1;i<=n;i++)
    {
        if(maxval<dis[i])maxval=dis[i];
    }
}
int main()
{
    ll T;
    scanf("%d",&T);
    while(T--)
    {
        int i,j,u,v,w;
        int V,E,S,K,C,id[maxn][maxn],k[maxn];
        memset(id,-1,sizeof(id));
        ll ans1,ans2;
        scanf("%d%d%d%d%d",&V,&E,&S,&K,&C);;
        for(i=1;i<=V+1;i++)p[i].clear();
        for(i=1;i<=K;i++)scanf("%d",&k[i]);
        for(i=1;i<=E;i++)
        {
            scanf("%d%d%d",&u,&v,&w);
            if(id[u][v]==-1)
            {
                id[u][v]=p[u].size();p[u].push_back(P(v,w));
                id[v][u]=p[v].size();p[v].push_back(P(u,w));
                continue;
            }
            if(p[u][id[u][v]].v<=w)continue;
            p[u][id[u][v]].v=w;p[v][id[v][u]].v=w;
        }
        n=V;
        Dij(S);ans1=maxval;
        n=V+1;
        for(i=1;i<=K;i++)
        {
            p[V+1].push_back(P(k[i],0));
            p[k[i]].push_back(P(V+1,0));
        }
        Dij(V+1);ans2=maxval;
//        printf("##%lld %lld\n",ans1,ans2);
        if(ans1<=ans2*C)printf("%lld\n",ans1);
        else printf("%lld\n",ans2);
    }
}