HDU

题意:给你一张带权无向图,先求出这张图从点1出发的最短路树,再求在树上经过k个节点最长的路径值,以及个数.
分析:首先求最短路树,跑一遍最短路之后dfs一遍即可建出最短路树.
第二个问题,树分治解决.
对于以root为根的树,所求的路径只会有两种情况.

1. 存在于root的子树中,不经过root;
2. 经过root,路径的两端在root的两棵子树中.
   第一种情况,我们交给分治去解决,
   第二种情况,需要知道所有子树中走过j步能到达的最远距离,以及其方案数.通过dfs可以得到这些信息.用这些信息,再去和其他子树的信息结合,去更新答案.

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int INF = 1<<30;
const int MAXN = 1e5+5;
struct Edge{
    int v,w,next;
}E[MAXN<<2];
int head[MAXN],tot , son[MAXN], Max[MAXN], siz[MAXN], dep[MAXN] ,now[MAXN];
LL cnt[MAXN], Maxcnt[MAXN] , ansnum;
int maxdep , clk, ansdep, minson;
bool vis[MAXN];
int root,N,M,k;

void init()
{
    memset(vis,0,sizeof(vis));
    memset(head,-1,sizeof(head));
    tot = ansnum = clk = 0;
    ansdep = 0;
}

void Add(int u,int v,int w){
    E[tot]  =(Edge){v,w,head[u]};
    head[u] = tot++;
}
//求子树的重心
void getsize(int u, int fa)
{
    siz[u] = 1;
    for (int i = head[u]; i != -1; i = E[i].next)
    {
        int v = E[i].v;
        if (v == fa || vis[v])
            continue;
        getsize(v, u);
        siz[u] += siz[v];
    }
}
void getroot(int u, int fa, int s)
{
    int max1 = 0;
    for (int i = head[u]; i != -1; i = E[i].next)
    {
        int v = E[i].v;
        if (v == fa || vis[v])
            continue;
        getroot(v, u, s);
        max1 = max(max1, siz[v]);
    }
    max1 = max(max1, s - siz[u]);
    if (minson > max1)
    {
        minson = max1;
        root = u;
    }
}
void getMaxdep(int depp, int u, int fa)
{
    maxdep = max(maxdep, depp);
    for (int i = head[u]; i != -1; i = E[i].next)
    {
        int v = E[i].v;
        if (v == fa || vis[v])
            continue;
        getMaxdep(depp + 1, v, u);
    }
}
void getdep(int depp, int len, int u, int fa)
{
    if (dep[depp] < len)
    {
        dep[depp] = len;
        cnt[depp] = 1;
    }
    else if (dep[depp] == len)
        cnt[depp]++;
    if (depp >= k)
        return;
    for (int i = head[u]; i != -1; i = E[i].next)
    {
        int v = E[i].v;
        if (v == fa || vis[v])
            continue;
        getdep(depp + 1, len + E[i].w, v, u);
    }
}
void getans(int u)
{
    vis[u] = 1;
    for (int i = head[u]; i != -1; i = E[i].next){
        int v = E[i].v;
        if (vis[v]) continue;
        minson = INF;
        getsize(v, -1);
        getroot(v, -1, siz[v]);
        getans(root);
    }
    /*
    求经过k个节点的最长路径,以及其方案数
    答案可能有两种情况, 一是存在于u的子树中,这种情况交给分治处理
    二是该路径经过了重心本身,在以下代码中处理
    */
    clk++;
    now[0] = clk;
    dep[0] = 0 ,cnt[0] = 1;
    Maxcnt[0] = 1, Max[0] = 0;
    for (int i = head[u]; i != -1; i = E[i].next){
        int v = E[i].v;
        if (vis[v])  continue;
        maxdep = -1;
        //获取这棵子树的最大深度
        getMaxdep(1, v, u);
        for (int j = 0; j <= maxdep && j <= k; j++)  dep[j] = -1;
        //获取这棵子树中经过i个节点,所能走的最长距离以及方案数,
        //分别记录在dep[i], 和cnt[i]中
        getdep(1, E[i].w, v, u);

        //根据当前信息和这棵子树的信息更新答案
        for (int j = 0; j <= maxdep && j < k; j++){
            int tmp = k - j - 1;
            if (now[tmp] != clk)
                continue;
            if (ansdep < Max[tmp] + dep[j]){
                ansdep = Max[tmp] + dep[j];
                ansnum = Maxcnt[tmp] * cnt[j];
            }
            else if (ansdep == Max[tmp] + dep[j]){
                ansnum += Maxcnt[tmp] * cnt[j];
            }
        }
        //用这个子树的信息更新当前的信息
        //Max[i]记录之前已经访问过的子树中,经过i个节点能走过的最长距离
        //Maxcnt[i] 记录其方案数
        for (int j = 1; j <= maxdep && j < k; j++){
            if (now[j] != clk || Max[j] < dep[j]){
                Max[j] = dep[j];
                Maxcnt[j] = cnt[j];
                now[j] = clk;
            }
            else if (now[j] == clk && Max[j] == dep[j]){
                Maxcnt[j] += cnt[j];
            }
        }
    }
    vis[u] = 0;
}

///////////最短路树
struct Dij{
    struct Edge{
        int v, w;
        bool operator < (const Edge & rhs) const{
            return v<rhs.v;
        }
    };
    vector<Edge> G[MAXN];
    int N,d[MAXN];
    bool vis[MAXN];

    struct HeapNode{
        int u,val;
        bool operator < (const HeapNode & rhs) const{
            return val>rhs.val;
        }
    };
    void init(int N){
        this -> N = N;
        memset(vis,0,sizeof(vis));
        for(int i=0;i<=N;++i) G[i].clear();
    }

    void AddEdge(int u,int v,int w){
        G[u].push_back((Edge){v,w});
    }

    void dijkstra(int s){
        for(int i=1;i<=N;++i){
            sort(G[i].begin(),G[i].end());
        }

        for(int i=0;i<=N;++i) d[i] = INF;
        d[s] = 0;
        priority_queue< HeapNode > Q;
        Q.push((HeapNode){s,0});
        while(!Q.empty()){
            HeapNode  x = Q.top(); Q.pop();
            if(vis[x.u]) continue;
            vis[x.u] = 1;
            int u = x.u, sz = G[u].size();
            for(auto & e : G[u]){
                int v = e.v;
                if(d[v]> d[u]+e.w){
                    d[v] = d[u] + e.w;
                    Q.push((HeapNode){v,d[v]});
                }
            }
        }
    }

    void dfs(int u,int fa){
        vis[u] = true;
        for(auto & e: G[u]){
            int v=  e.v;
            if(v==fa || vis[v]) continue;
            if(d[v]== d[u]+e.w){
                Add(u,v,e.w);
                Add(v,u,e.w);
                dfs(v,u);
            }
        }
    }
}G;

////////////////// main
int main()
{
    #ifndef ONLINE_JUDGE
        freopen("in.txt","r",stdin);
        freopen("out.txt","w",stdout);
    #endif
    int T; scanf("%d",&T);
    while(T--){
        scanf("%d %d %d",&N, &M, &k);
        G.init(N);
        int u,v,w;
        while(M--){
            scanf("%d %d %d",&u, &v, &w);
            G.AddEdge(u,v,w);
            G.AddEdge(v,u,w);
        }
        G.dijkstra(1);
        init();              //树的初始化
        memset(G.vis,0,sizeof(G.vis));
        G.dfs(1,-1);           //建最短路径树

        memset(now,-1,sizeof(now));
        root = -1;
        minson = INF;
        getroot(1,-1,N);
        getans(root);
        printf("%d %lld
",ansdep,ansnum);
    }
    return 0;
}