2017多校第4场 HDU 6074 Phone Call 并查集，LCA

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6074

题意：给你一棵树，然后给你M个条件，每次给出a,b,c,d,cost,表示从a-->b,c-->d的路径中的点，可以互相到达，花费是cost，到达具有传递性 ，现在问你从1节点最多可以到达哪些节点，最小花费是多少。

解法：看着官方题解学的。

#include <bits/stdc++.h>using namespace std;const int maxn=1e5+10;typedef long long LL;int T,n,m,parent[maxn],up[maxn],cnt[maxn];struct FastIO{    static const int S = 1310720;    int wpos;    char wbuf[S];    FastIO() : wpos(0) {}    inline int xchar()    {        static char buf[S];        static int len = 0, pos = 0;        if (pos == len)            pos = 0, len = fread(buf, 1, S, stdin);        if (pos == len) return -1;        return buf[pos ++];    }    inline LL Lint()    {        int c = xchar();        LL x = 0;        while (c <= 32) c = xchar();        for (; '0' <= c && c <= '9'; c = xchar()) x = x * 10 + c - '0';        return x;    }    inline int xint()    {        int s = 1, c = xchar(), x = 0;        while (c <= 32) c = xchar();        if (c == '-') s = -1, c = xchar();        for (; '0' <= c && c <= '9'; c = xchar()) x = x * 10 + c - '0';        return x * s;    }    inline void xstring(char *s)    {        int c = xchar();        while (c <= 32) c = xchar();        for (; c > 32; c = xchar()) * s++ = c;        *s = 0;    }    inline void wchar(int x)    {        if (wpos == S) fwrite(wbuf, 1, S, stdout), wpos = 0;        wbuf[wpos ++] = x;    }    inline void wint(LL x)    {        if (x < 0) wchar('-'), x = -x;        char s[24];        int n = 0;        while (x || !n) s[n ++] = '0' + x % 10, x /= 10;        while (n--) wchar(s[n]);    }    inline void wstring(const char *s)    {        while (*s) wchar(*s++);    }    ~FastIO()    {        if (wpos) fwrite(wbuf, 1, wpos, stdout), wpos = 0;    }} io;//parent[i]表示父亲节点是谁//up[i]表示i点往上深度最大的一个可能不是和i在同一个连通块的祖先//cnt[i]表示i所在连通块的点数LL w[maxn]; //i所在连通块离的距离和int dep[maxn],fa[maxn][20];//dep[i]表示i点的深度,fa[i][j]代表i跳2^j的点vector <int> G[maxn];struct node{    int a,b,c,d;    LL cost;    bool operator < (const node &rhs) const{        return cost < rhs.cost;    }}q[maxn];//存下询问,按v值排序void dfs1(int x, int d, int pre){    dep[x] = d;    fa[x][0] = pre;    for(int i=1; i<20; i++){        fa[x][i] = fa[fa[x][i-1]][i-1];    }    for(int i=0; i<G[x].size(); i++){        int v=G[x][i];        if(v == pre) continue;        dfs1(v, d+1, x);    }}int findfa(int x){    if(x==parent[x]) return x;    else return parent[x] = findfa(parent[x]);}int findup(int x){    if(x == up[x]) return x;    else return up[x] = findup(up[x]);}int LCA(int u, int v){    if(dep[u]<dep[v]) swap(u,v);    for(int i=19; i>=0; i--){        if(dep[fa[u][i]]>=dep[v]){            u=fa[u][i];        }    }    if(u==v) return u;    for(int i=19; i>=0; i--){        if(fa[u][i]!=fa[v][i]){            u=fa[u][i];            v=fa[v][i];        }    }    return fa[u][0];}void Union(int u, int v, LL cost){    u = findfa(u), v = findfa(v);    if(u == v) return;    parent[u] = v;    cnt[v] += cnt[u];    w[v] += w[u] + cost;}void Merge(int u, int v, LL cost){    while(1){        u=findup(u);        if(dep[u]<=dep[v]) return;        Union(u, fa[u][0], cost);        up[u] = fa[u][0];    }}void solve(node s){    int _lca = LCA(s.a, s.b);    Merge(s.a, _lca, s.cost);    Merge(s.b, _lca, s.cost);    _lca = LCA(s.c, s.d);    Merge(s.c, _lca, s.cost);    Merge(s.d, _lca, s.cost);    Union(s.a, s.c, s.cost);}int main(){    //freopen("1.in","r",stdin);    //freopen("1.out","w",stdout);    T = io.xint();    while(T--)    {        n = io.xint(), m = io.xint();        for(int i=0; i<=n; i++) parent[i]=up[i]=i,cnt[i]=1,w[i]=0,G[i].clear();        for(int i=1; i<n; i++){            int u, v;            u = io.xint();            v = io.xint();            G[u].push_back(v);            G[v].push_back(u);        }        for(int i=1; i<=m; i++){            //scanf("%d %d %d %d %lld", &q[i].a, &q[i].b, &q[i].c, &q[i].d, &q[i].cost);            q[i].a = io.xint();            q[i].b = io.xint();            q[i].c = io.xint();            q[i].d = io.xint();            q[i].cost = io.Lint();        }        sort(q+1, q+m+1);        dfs1(1, 1, 0);        for(int i=1; i<=m; i++)        {            solve(q[i]);        }        printf("%d %lld\n", cnt[findfa(1)], w[findfa(1)]);    }    return 0;}