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  • poj 1986 Distance Queries

    LCA

    题意:LCA模板题,输入n和m,表示n个点m条边,下面m行是边的信息,两端点和权,后面的那个字母无视掉,没用的。接着k,下面k个询问lca,输出即可

    有人说要考虑不连通的情况,我没考虑AC了,另外可能有u,u这样的询问,不过这不影响,照样是写模板,没有特判,一样能过

    还是Tarjan快一些

    LCA转RMQ在线算法

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <cmath>
    using namespace std;
    #define N 40010
    #define M 25
    
    int tot;
    int __pow[M];
    int head[N];
    struct edge{
        int u,v,w,next;
    }e[2*N];
    int ver[2*N],R[2*N],first[N],dir[N];
    int dp[2*N][25];
    bool vis[N];
    
    inline void add(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++;
        u = u^v; v = u^v; u = u^v;
        e[k].u = u; e[k].v = v; e[k].w = w;
        e[k].next = head[u]; head[u] = k++;
    }
    
    void dfs(int u ,int dep)
    {
        vis[u] = true; first[u] = ++tot; ver[tot] = u; R[tot] = dep;
        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;
                dfs(v,dep+1);
                ver[++tot] = u; R[tot] = dep;
            }
    }
    
    void ST(int len)
    {
        int K = (int)(log((double)(len)) / log(2.0));
        for(int i=1; i<=len; i++) dp[i][0] = i;
        for(int j=1; j<=K; j++)
            for(int i=1; i+__pow[j]-1 <= len; i++)
            {
                int a = dp[i][j-1] , b = dp[i+__pow[j-1]][j-1];
                if(R[a] < R[b]) dp[i][j] = a;
                else            dp[i][j] = b;
            }
    }
    
    int RMQ(int x ,int y)
    {
        int K = (int)(log((double)(y-x+1)) / log(2.0));
        int a = dp[x][K] , b = dp[y-__pow[K]+1][K];
        if(R[a] < R[b]) return a;
        else            return b;
    }
    
    int LCA(int u ,int v)
    {
        int x = first[u] , y = first[v];
        if(x > y) swap(x,y);
        int index = RMQ(x,y);
        return ver[index];
    }
    
    int main()
    {
        for(int i=0; i<M; i++) __pow[i] = (1<<i);
        int n,m,k,str[10];
        while(scanf("%d%d",&n,&m)!=EOF)
        {
            k = 0;
            memset(head,-1,sizeof(head));
            memset(vis,false,sizeof(vis));
            while(m--)
            {
                int u,v,w;
                scanf("%d%d%d%s",&u,&v,&w,str);
                add(u,v,w,k);
            }
            tot = dir[1] = 0;
            dfs(1,1);
            ST(tot);
            int q;
            scanf("%d",&q);
            while(q--)
            {
                int u,v,lca;
                scanf("%d%d",&u,&v);
                lca = LCA(u,v);
                printf("%d\n",dir[u] + dir[v] - 2*dir[lca]);
            }
        }
        return 0;
    }

    Tarjan离线算法

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    using namespace std;
    #define N 40010
    #define M 20010
    
    int head[N];
    struct edge{
        int u,v,w,next;
    }e[2*N];
    int __head[N];
    struct ask{
        int u,v,lca,next;
    }ea[M];
    int fa[N],ance[N],dir[N];
    bool 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++;
        u = u^v; v = u^v; u = u^v;
        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++;
        u = u^v; v = u^v; u = u^v;
        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)
    {
        vis[u] = true;
        ance[u] = 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);
                fa[v] = u;
            }
        for(int k=__head[u]; k!=-1; k=ea[k].next)
            if( vis[ea[k].v] )
                ea[k].lca = ea[k^1].lca = ance[find(ea[k].v)];
    }
    
    int main()
    {
        int n,m,q,k; char str[10];
        while(scanf("%d%d",&n,&m)!=EOF)
        {
            memset(head,-1,sizeof(head));
            memset(__head,-1,sizeof(__head));
            memset(vis,false,sizeof(vis));
            k = 0;
            while(m--)
            {
                int u,v,w;
                scanf("%d%d%d%s",&u,&v,&w,str);
                add_edge(u,v,w,k);
            }
            scanf("%d",&q);
            k = 0;
            for(int i=0; i<q; i++)
            {
                int u ,v;
                scanf("%d%d",&u,&v);
                add_ask(u,v,k);
            }
            dir[1] = 0;
            Tarjan(1);
            for(int i=0; i<q; i++)
            {
                int s = i*2 , u = ea[s].u , v = ea[s].v , lca = ea[s].lca;
                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/3102068.html
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