FJUT 聪明的商人（树上倍增）题解

思路：求树上两点的距离，显然是dep[u] + dep[v] - 2 * dep[lca]，用树上倍增去写。

参考：树上倍增的写法和应用（详细讲解，新手秒懂）

代码：

#include<set>
#include<map>
#include<stack>
#include<cmath>
#include<queue>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
typedef long long ll;
const int maxn =  + ;
const int seed = ;
const ll MOD = 1e9 + ;
const ll INF = 1e17;
using namespace std;
struct Edge{
    int to, next;
}edge[maxn << ];
int fa[maxn][], dep[maxn], head[maxn], tot, maxf;
void addEdge(int u, int v){
    edge[tot].to = v;
    edge[tot].next = head[u];
    head[u] = tot++;
}
void dfs(int x, int pre, int deep){
    dep[x] = deep;
    fa[x][] = pre;
    for(int i = ; i <= maxf; i++){
        if(fa[x][i - ] != -){
            fa[x][i] = fa[fa[x][i - ]][i - ];
        }
        else break;
    }
    for(int i = head[x]; i != -; i = edge[i].next){
        int v = edge[i].to;
        if(v != pre){
            dfs(v, x, deep + );
        }
    }
}
int LCA(int u, int v){
    if(dep[u] < dep[v]) swap(u, v);
    int dis = dep[u] - dep[v];
    for(int i = ; i <= maxf; i++){
        if(( << i) & dis)
            u = fa[u][i];
    }
    if(u == v) return u;
    for(int i = maxf; i >= ; i--){
        if(fa[u][i] != fa[v][i]){
            u = fa[u][i];
            v = fa[v][i];
        }
    }
    return fa[u][];
}

void init(){
    maxf = ;
    memset(head, -, sizeof(head));
    memset(fa, -, sizeof(fa));
    tot = ;
}
int main(){
    int n, m, w;

    while(~scanf("%d%d%d", &n, &m, &w)){
        init();
        for(int i = ; i < n - ; i++){
            int a, b;
            scanf("%d%d", &a, &b);
            addEdge(a, b);
            addEdge(b, a);
        }
        dfs(, -, );
        for(int i = ; i < m; i++){
            int p, q, v;
            scanf("%d%d%d", &p, &q, &v);
            int lca = LCA(p , q);
            int dis = dep[p] + dep[q] -  * dep[lca];
            if(v - dis * w >= ){
                printf("%d\n", v - dis * w);
            }
            else{
                printf("pass\n");
            }
        }
    }
    return ;
}