BZOJ1602: [Usaco2008 Oct]牧场行走

题目传送门

题解：一个简单LCA即可 记录一下根到每个节点的距离

然后对于每个询问（a，b）的答案就是根到a的距离+根到b的距离-2*根到LCA（a，b）的距离

#include<bits/stdc++.h>#define ll long longconst int INF = 0x7fffffff;const double eps = 1e-5;const int maxn = 1e3 + 5;using namespace std;int read(){int x = 0 , f = 1; char ch = getchar();while(ch<'0'||ch>'9') {if(ch=='-')f*=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+(ch-'0');ch=getchar();}return x * f;}int n,q;int cnt,to[maxn<<1],nxt[maxn<<1],c[maxn<<1],head[maxn];int deep[maxn],fa[maxn][18],dis[maxn],vis[maxn];void add(int u,int v,int w){to[++cnt] = v ; nxt[cnt] = head[u] ; head[u] = cnt; c[cnt] = w;to[++cnt] = u ; nxt[cnt] = head[v] ; head[v] = cnt; c[cnt] = w;}void dfs(int x,int f){vis[x]=1;for(int i = 1; i <= 9; i++){if(deep[x]<(1<<i)) break;fa[x][i] = fa[fa[x][i-1]][i-1];}for(int i = head[x]; i; i = nxt[i]){if(to[i] != f){deep[to[i]] = deep[x]+1;dis[to[i]] = dis[x] + c[i];fa[to[i]][0]=x;dfs(to[i],x);}}}int LCA(int a,int b){if(deep[b] > deep[a]) swap(a,b);int t = deep[a] - deep[b];for(int i = 0 ; i <= 9 ; i++)if(t&(1<<i)) a = fa[a][i];for(int i = 0 ; i <= 9 ; i++){while(fa[a][i]!=fa[b][i])a=fa[a][i],b=fa[b][i];}if(a==b) return a;else return fa[a][0];}int main(){n = read() , q = read();for(int i = 1 ; i < n ; i++){int a = read() , b = read() , c = read();add(a,b,c);}dfs(1,0);for(int i = 1 ; i <= q ; i++){int a = read(), b =read();printf("%d\n",dis[a]+dis[b]-(dis[LCA(a,b)]<<1));}return 0;}