BZOJ 4326 运输计划

二分答案+树链剖分+树上差分

我们假设x是最小的花费，可以想到给定x，所有运输计划中花费大于x的计划必须经过虫洞，且最长的一条的花费减去虫洞所在边的花费要小于等于x
那么对于x，虫洞所在的位置肯定是确定的，假设x可以取更小，那么就没有一个合法方案可以放虫洞，x取更大，显然该方案也合法，这是一个明显符合单调性的问题，我们可以用二分答案求解。
其实最大值最小就是答案具有单调性的特征啦。。

通过上述分析，我们可以确定虫洞所在位置就是花费大于x的运输计划的交，即该边被覆盖次数等于花费大于x的运输计划数，那么对于每个x，我们可以用树上边差分的方式来求出交。
如何求最大的花费呢？我们考虑差分时要求LCA，且要求出树上的路径，那么树链剖分就再好不过了！
在dfs1的时候我们就需要吧边权塞给点（便是该点到其父亲的权值），然后通过树链剖分把LCA，和两点花费求出来。最后用max维护最大即可。

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define full(a, b) memset(a, b, sizeof a)
using namespace std;
typedef long long ll;
inline int lowbit(int x){ return x & (-x); }
inline int read(){
    int X = 0, w = 0; char ch = 0;
    while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }
    while(isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
    return w ? -X : X;
}
inline int gcd(int a, int b){ return a % b ? gcd(b, a % b) : b; }
inline int lcm(int a, int b){ return a / gcd(a, b) * b; }
template<typename T>
inline T max(T x, T y, T z){ return max(max(x, y), z); }
template<typename T>
inline T min(T x, T y, T z){ return min(min(x, y), z); }
template<typename A, typename B, typename C>
inline A fpow(A x, B p, C lyd){
    A ans = 1;
    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;
    return ans;
}
const int N = 300005;
int n, m, cnt, dfn, head[N], size[N], p[N], son[N], depth[N], top[N], w[N], val[N], id[N], tmp[N];
int tree[N<<2], num, len, ml, ans;
struct Edge { int v, next, w; } edge[N<<1];
struct Ask {
    int u, v, lca, dis;
    bool operator < (const Ask &rhs) const {
        return dis > rhs.dis;
    }
} ask[N<<1];

void addEdge(int a, int b, int w){
    edge[cnt].v = b, edge[cnt].w = w, edge[cnt].next = head[a], head[a] = cnt ++;
}

void dfs1(int s, int fa){
    depth[s] = depth[fa] + 1;
    p[s] = fa;
    size[s] = 1;
    int child = -1;
    for(int i = head[s]; i != -1; i = edge[i].next){
        int u = edge[i].v;
        if(u == fa) continue;
        dfs1(u, s);
        size[s] += size[u], val[u] = edge[i].w;
        if(size[u] > child) child = size[u], son[s] = u;
    }
}

void dfs2(int s, int tp){
    id[s] = ++dfn;
    w[id[s]] = val[s];
    top[s] = tp;
    if(son[s] != -1) dfs2(son[s], tp);
    for(int i = head[s]; i != -1; i = edge[i].next){
        int u = edge[i].v;
        if(u == p[s] || u == son[s]) continue;
        dfs2(u, u);
    }
}

void push_up(int rt){
    tree[rt] = tree[rt << 1] + tree[rt << 1 | 1];
}

void buildTree(int rt, int l, int r){
    if(l == r){
        tree[rt] = w[l];
        return;
    }
    int mid = (l + r) >> 1;
    buildTree(rt << 1, l, mid);
    buildTree(rt << 1 | 1, mid + 1, r);
    push_up(rt);
}

int query(int rt, int l, int r, int queryL, int queryR){
    if(queryL > queryR) return 0;
    if(l == queryL && r == queryR){
        return tree[rt];
    }
    int mid = (l + r) >> 1;
    if(queryL > mid) return query(rt << 1 | 1, mid + 1, r, queryL, queryR);
    else if(queryR <= mid) return query(rt << 1, l, mid, queryL, queryR);
    else return query(rt << 1, l, mid, queryL, mid) +
                query(rt << 1 | 1, mid + 1, r, mid + 1, queryR);
}

void lca(int x, int y, int k){
    int v = 0;
    while(top[x] != top[y]){
        if(depth[top[x]] < depth[top[y]]) swap(x, y);
        v += query(1, 1, n, id[top[x]], id[x]);
        x = p[top[x]];
    }
    if(depth[x] > depth[y]) swap(x, y);
    v += query(1, 1, n, id[x] + 1, id[y]);
    ask[k].lca = x, ask[k].dis = v;
}

void dfs3(int s, int fa){
    for(int i = head[s]; i != -1; i = edge[i].next){
        int u = edge[i].v;
        if(u == fa) continue;
        dfs3(u, s);
        tmp[s] += tmp[u];
    }
    if(s != 1 && tmp[s] == num && val[s] > len) len = val[s];
}

bool check(int x){
    num = 0, len = -INF, ml = -INF;
    full(tmp, 0);
    int i = 0;
    for(; i < m; i ++){
        if(ask[i].dis <= x) break;
        int u = ask[i].u, v = ask[i].v, f = ask[i].lca;
        tmp[u] ++, tmp[v] ++, tmp[f] -= 2;
        num ++;
        ml = max(ml, ask[i].dis);
    }
    if(i == 0) return true;
    dfs3(1, 0);
    return ml - len <= x;
}

void solve(){
    sort(ask, ask + m);
    int l = 0, r = ask[0].dis;
    while(l < r){
        int mid = (l + r) >> 1;
        if(check(mid)) r = mid;
        else l = mid + 1;
    }
    ans = l;
}

int main(){

    full(head, -1);
    full(son, -1);
    n = read(), m = read();
    for(int i = 0; i < n - 1; i ++){
        int u = read(), v = read(), w = read();
        addEdge(u, v, w), addEdge(v, u, w);
    }
    dfs1(1, 0), dfs2(1, 1);
    buildTree(1, 1, n);
    for(int i = 0; i < m; i ++){
        ask[i].u = read(), ask[i].v = read();
        //cout << i << " " << ask[i].u << " " << ask[i].v << endl;
        lca(ask[i].u, ask[i].v, i);
    }
    solve();
    printf("%d\n", ans);
    return 0;
}