WOJ 43 电话邀请

并查集缩点这个trick感觉明明用得很广泛，为什么以前都不知道……

先把$m$条线路从小到大排个序，这样可以保证之前合并出来的一定是最小的，大的代价不会把小的覆盖掉。

维护两个并查集，一个用来缩点，另一个用来维护生成树的相关信息

直接把每一条树链合并到lca处，最后再把两个lca合并，因为最后要把两个lca合并，所以求lca拆开跳链的做法比较优秀。

链剖求lca还真的比倍增常数小

感觉get了很多。

Code：

#include <cstdio>
#include <algorithm>
using namespace std;
typedef long long ll;

const int N = 1e5 + 5;
const int Lg = 20;

int n, m, tot = 0, head[N], ufs[N];
int fa[N][Lg], same[N], dep[N], siz[N];
ll cost[N];

struct Edge {
    int to, nxt;
} e[N << 1];

inline void add(int from, int to) {
    e[++tot].to = to;
    e[tot].nxt = head[from];
    head[from] = tot;
}

struct Lineway {
    int u1, v1, u2, v2;
    ll cost; 
} a[N];

bool cmp(const Lineway &x, const Lineway &y) {
    return x.cost < y.cost;
}

template <typename T>
inline void read(T &X) {
    X = 0;
    char ch = 0;
    T op = 1;
    for(; ch > '9'|| ch < '0'; ch = getchar())
        if(ch == '-') op = -1;
    for(; ch >= '0' && ch <= '9'; ch = getchar())
        X = (X << 3) + (X << 1) + ch - 48;
    X *= op;
}

inline void swap(int &x, int &y) {
    int t = x;
    x = y;
    y = t;
}

void dfs(int x, int fat, int depth) {
    fa[x][0] = fat, dep[x] = depth;
    for(int i = 1; i <= 18; i++)
        fa[x][i] = fa[fa[x][i - 1]][i - 1];
    for(int i = head[x]; i; i = e[i].nxt) {
        int y = e[i].to;
        if(y == fat) continue;
        dfs(y, x, depth + 1);
    }
}

inline int getLca(int x, int y) {
    if(dep[x] < dep[y]) swap(x, y);
    for(int i = 18; i >= 0; i--)
        if(dep[fa[x][i]] >= dep[y])
            x = fa[x][i];
    if(x == y) return x;
    for(int i = 18; i >= 0; i--)
        if(fa[x][i] != fa[y][i])    
            x = fa[x][i], y = fa[y][i];
    return fa[x][0];
}

inline void init() {
    for(int i = 1; i <= n; i++)
        same[i] = i, ufs[i] = i, siz[i] = 1, cost[i] = 0LL;
}

int find(int x) {
    return ufs[x] == x ? x : ufs[x] = find(ufs[x]); 
}

int findSame(int x) {
    return same[x] == x ? x : same[x] = findSame(same[x]);
}

inline void merge(int x, int y, ll c) {
    int fx = find(x), fy = find(y);
    if(fx == fy) return;
    ufs[fx] = fy;
    siz[fy] += siz[fx];
    cost[fy] += cost[fx] + c;
}

inline void go(int x, int y, ll c) {
    for(; ; ) {
        x = findSame(x);
        if(dep[x] <= dep[y]) return;
        merge(x, fa[x][0], c);
        same[x] = fa[x][0];
    }
}

inline void chain(int x, int y, ll c) {
    int z = getLca(x, y);
    go(x, z, c), go(y, z, c);
}

int main() {
    read(n), read(m);
    for(int x, y, i = 1; i < n; i++) {
        read(x), read(y);
        add(x, y), add(y, x);
    }
    dfs(1, 0, 1);
    
    for(int i = 1; i <= m; i++)
        read(a[i].u1), read(a[i].v1), read(a[i].u2), read(a[i].v2), read(a[i].cost);
    sort(a + 1, a + 1 + m, cmp);
    
    init();
    for(int i = 1; i <= m; i++) {
        chain(a[i].u1, a[i].v1, a[i].cost);
        chain(a[i].u2, a[i].v2, a[i].cost);
        merge(a[i].u1, a[i].u2, a[i].cost);
    }
    
    int ans = find(1);
    printf("%d %lld
", siz[ans], cost[ans]);
    return 0;
}