Newcoder Metropolis（多源最短路 + Dijkstra堆优化）题解

题目链接：https://www.nowcoder.com/acm/contest/203/I?tdsourcetag=s_pcqq_aiomsg
来源：牛客网

思路：我们用用fa[i]表示距离i最近的大都市，dis[i]表示i距离该大都市的距离。我们先把所有大都市加入初始点，然后跑Dijkstra，如果某一点到另一个大都市距离更近，那么更新dis和fa。如果取边时，边的两边分别属于不同的两个大都市，说明两个大都市连通了，那么更新答案。

代码：

#include<set>
#include<map>
#include<stack>
#include<cmath>
#include<queue>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
typedef long long ll;
const int maxn =  + ;
const int seed = ;
const ll MOD = 1e9 + ;
const ll INF = 1e17;
using namespace std;
struct Edge{
    int to, next;
    ll w;
}edge[maxn << ];
struct qnode{
    int u;
    ll c;
    qnode(int _u = , ll _c = ):u(_u), c(_c){}
    bool operator < (const qnode &r) const{
        return r.c < c;
    }
};
int n, m, p, tot, head[maxn], big[maxn], fa[maxn], id[maxn];
ll dis[maxn], ans[maxn];
void addEdge(int u, int v, ll w){
    edge[tot].to = v;
    edge[tot].w = w;
    edge[tot].next = head[u];
    head[u] = tot++;
}
void Dijkstra(){
    memset(fa, -, sizeof(fa));
    for(int i = ; i <= n; i++) dis[i] = INF;
    for(int i = ; i < p; i++) ans[i] = INF;
    priority_queue<qnode> que;
    while(!que.empty()) que.pop();
    for(int i = ; i < p; i++){
        dis[big[i]] = ;
        fa[big[i]] = big[i];
        que.push(qnode(big[i], ));
    }
    qnode temp;
    while(!que.empty()){
        temp = que.top();
        que.pop();
        int u = temp.u;
        for(int i = head[u]; i != -; i = edge[i].next){
            int v = edge[i].to;
            ll w = edge[i].w;
            if(dis[v] > dis[u] + w){
                dis[v] = dis[u] + w;
                fa[v] = fa[u];
                que.push(qnode(v, dis[v]));
            }
            else if(fa[u] != fa[v] && fa[v] != -){
                ans[id[fa[u]]] = min(ans[id[fa[u]]], dis[u] + dis[v] + w);
                ans[id[fa[v]]] = min(ans[id[fa[v]]], dis[u] + dis[v] + w);
            }
        }
    }
}
int main(){
    int u, v;
    ll w;
    scanf("%d%d%d", &n, &m, &p);
    memset(head, -, sizeof(head));
    tot = ;
    for(int i = ; i < p; i++){
        scanf("%d", &u);
        big[i] = u;
        id[u] = i;
    }
    for(int i = ; i < m; i++){
        scanf("%d%d%lld", &u, &v, &w);
        addEdge(u, v, w);
        addEdge(v, u, w);
    }
    Dijkstra();
    for(int i = ; i < p; i++){
        if(i != ) printf(" ");
        printf("%lld", ans[i]);
    }
    printf("\n");
    return ;
}