牛客国庆集训派对Day3 I.

题意:求一个N个点无向图中,其中p个关键点间的最短距离.
分析:比较特殊的最短路,方式类似于多源BFS,将所有关键点装入优先队列,状态中需要包含其源点的id.对每条边都要遍历,对每个节点,需要记录其确定最短的源头以及其最短距离.当一个访问状态到达了与自己源头状态不同的点,则说明两个关键点相遇,每次相遇时,更新两个源头的最短距离.

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int MAXN = 2e5+5;
const LL INF = (1LL)<<60;
struct Edge{
    int v,next;
    LL w;
}E[MAXN<<2];
int head[MAXN],tot;
int vis[MAXN];
LL d[MAXN],link[MAXN];
vector<int> st;
int N,M,k;

void init()
{
    st.clear();
    memset(head,-1,sizeof(head));
    tot = 0;
}

void AddEdge(int u,int v,int w){
    E[tot] = (Edge){v,head[u],w};
    head[u] = tot++;
}

struct HeapNode{
    int u,sta;
    LL val;
    bool operator < (const HeapNode & rhs) const{
        return val > rhs.val;
    }
};
void Dijkstra()
{
    for(int i=0;i<=N;++i)   d[i] = INF, vis[i] = 0;

    priority_queue<HeapNode> Q;
    for(int i=0,sz = st.size();i<sz;++i){
        int u = st[i];
        d[u] = 0;
        link[u] = INF;
        Q.push((HeapNode){u,u,0});
    }

    while(!Q.empty()){
        HeapNode x = Q.top(); Q.pop();
        int u = x.u, sta = x.sta;
        if(vis[u] == 0){
            vis[u] = sta;
            d[u] = x.val;
            for(int i=head[u]; ~i; i = E[i].next){
                int v = E[i].v;
                Q.push((HeapNode){v,sta,d[u]+E[i].w});
            }
        }

        else if(vis[u] != sta){
            int fu = vis[u];
            link[sta] = min(link[sta], x.val + d[u]);
            link[fu] = min(link[fu], x.val+ d[u]);
        }
    }
}

int main()
{
    #ifndef ONLINE_JUDGE
        freopen("in.txt","r",stdin);
        freopen("out.txt","w",stdout);
    #endif
    scanf("%d %d %d",&N, &M, &k);
    init();
    int u,v;
    LL w;
    while(k--){
        scanf("%d",&u);
        st.push_back(u);
    }
    while(M--){
        scanf("%d %d %lld",&u,&v,&w);
        AddEdge(u,v,w);
        AddEdge(v,u,w);
    }
    Dijkstra();
    for(int i=0,sz = st.size();i<sz;++i){
        printf("%lld%c",link[st[i]], i==sz-1?'
':' ');
    }
    return 0;
}