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  • hdu 2874 Connections between cities

    LCA

    题意:询问LCA,不过图不连通,如果两点不连通输出那串英文否则输出两点间的距离,模板题

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    using namespace std;
    #define N 10010
    #define M 10010
    #define Q 1000010
    
    int head[N];
    struct edge{
        int u,v,w,next;
    }e[2*M];
    int __head[N];
    struct ask{
        int u,v,lca,next;
    }ea[2*Q];
    int dir[N],fa[N],vis[N];
    
    inline void add_edge(int u ,int v ,int w ,int &k)
    {
        e[k].u = u; e[k].v = v; e[k].w = w;
        e[k].next = head[u]; head[u] = k++;
    }
    inline void add_ask(int u ,int v ,int &k)
    {
        ea[k].u = u; ea[k].v = v; ea[k].lca = -1;
        ea[k].next = __head[u]; __head[u] = k++;
    }
    
    int find(int x){
        return  x == fa[x] ? x : fa[x] = find(fa[x]);
    }
    
    void Tarjan(int u ,int c)
    {
        vis[u] = c; fa[u] = u;
        for(int k=head[u]; k!=-1; k=e[k].next)
            if(!vis[e[k].v])
            {
                int v = e[k].v , w = e[k].w;
                dir[v] = dir[u] + w;
                Tarjan(v,c);
                fa[v] = u;
            }
        for(int k=__head[u]; k!=-1; k=ea[k].next)
            if(vis[ea[k].v] == c)
                ea[k].lca = ea[k^1].lca = find(ea[k].v);
    }
    
    int main()
    {
        int n,m,q,k;
        while(scanf("%d%d%d",&n,&m,&q)!=EOF)
        {
            memset(head,-1,sizeof(head));
            memset(__head,-1,sizeof(__head));
            memset(vis,0,sizeof(vis));
            k = 0;
            for(int i=0; i<m; i++)
            {
                int u,v,w;
                scanf("%d%d%d",&u,&v,&w);
                add_edge(u,v,w,k);
                add_edge(v,u,w,k);
            }
            k = 0;
            for(int i=0; i<q; i++)
            {
                int u,v;
                scanf("%d%d",&u,&v);
                add_ask(u,v,k);
                add_ask(v,u,k);
            }
            k = 0;
            for(int i=1; i<=n; i++)
                if(!vis[i])
                {
                    dir[i] = 0;
                    Tarjan(i,++k);
                }
            for(int i=0; i<q; i++)
            {
                int s = i*2 , u = ea[s].u , v = ea[s].v , lca = ea[s].lca;
                if(lca == -1) printf("Not connected\n");
                else          printf("%d\n",dir[u] + dir[v] - 2*dir[lca]);
            }
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/scau20110726/p/3107462.html
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