hdu 2586 How far away ？[LCA/ST+RMQ]

题目链接

题意：

给你一棵树，让你求出树上的两个点之间的距离。

思路：

用LCA解决，求出两个节点到根节点的距离减去两倍的LCA到根节点的距离

#include "bits/stdc++.h"
using namespace std;  
const int maxn = 40010;   
int dp[2*maxn][30];   
bool vis[maxn];  
int ver[2*maxn], dept[2*maxn];
int first[maxn], dis[maxn];  
struct edge {
    int u,v,w,next;  
} e[2*maxn];  
int cnt,head[maxn];  
int num;
void add(int u ,int v ,int w) {
    e[num].u = u; e[num].v = v; e[num].w = w;  
    e[num].next = head[u]; head[u] = num++;  
}   
void dfs(int u ,int dep) {
    vis[u] = true; ver[++cnt] = u; first[u] = cnt; dept[cnt] = dep;  
    for(int i = head[u]; i != -1; i = e[i].next) { 
        if(!vis[e[i].v]) {
            int v = e[i].v , w = e[i].w;  
            dis[v] = dis[u] + w;  
            dfs(v,dep+1);  
            ver[++cnt] = u; dept[cnt] = dep;  
        }
    }          
}  
void ST(int n) {
    for(int i = 1; i <= n; i++) dp[i][0] = i;  
    for(int j = 1; (1<<j) <= n; j++) {
        for(int i = 1; i+(1<<j)-1 <= n; i++) {
            int a = dp[i][j-1], b = dp[i+(1<<(j-1))][j-1];  
            dp[i][j] = dept[a] < dept[b]?a:b;  
        }  
    }  
}  
int RMQ(int l,int r) {
    int k=0;  
    while((1<<(k+1)) <= r-l+1) k++;  
    int a = dp[l][k], b = dp[r-(1<<k)+1][k];   
    return dept[a] < dept[b]?a:b;  
}  
int LCA(int u ,int v) {
    int x = first[u] , y = first[v];  
    if(x > y) swap(x,y);  
    int res = RMQ(x,y);  
    return ver[res];  
}  
int main(int argc, char const *argv[])
{
    int T;  
    scanf("%d", &T);  
    while(T--) {
        int n, q;
        num = 0;  
        scanf("%d%d", &n, &q);  
        memset(head, -1, sizeof(head));  
        memset(vis, false, sizeof(vis));  
        for(int i=1; i<n; i++) {
            int u,v,w;  
            scanf("%d%d%d",&u,&v,&w);  
            add(u,v,w);  
            add(v,u,w);
        }  
        cnt = 0; dis[1] = 0;  
        dfs(1,1);
        ST(2*n-1);  
        while(q--) {
            int u,v;  
            scanf("%d%d",&u,&v);  
            int lca = LCA(u,v);  
            printf("%d\n",dis[u] + dis[v] - 2*dis[lca]);  
        }  
    }  
    return 0;  
}