zoj3195(lca / RMQ在线)

题目链接: http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3195

题意: 给出一棵 n 个节点的带边权的树, 有 q 组形如 x, y, z 的询问, 输出 x, y, z之间的最短路径.

思路: 在纸上画下不难发现 x, y, z之间的最短路径就是 x, y, z 两两之间的最短路径和的一半.

我们可以通过 lca 模板求出 x, y, z 两两之间的最短路径, 然后再算下 x, y, z三点之间的最短路径即可.

这题应该是用 RMQ 在线比较好写一点, 用 tarjan 的话记录路径有点麻烦.

代码:

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <math.h>
using namespace std;

const int MAXN = 5e4 + 10;
struct node{
    int v, w, next;
}edge[MAXN << 1];

int dp[MAXN << 1][30];
int ver[MAXN << 1], deep[MAXN << 1], first[MAXN];
int dis[MAXN], head[MAXN], vis[MAXN], indx, ip;

inline void init(void){
    memset(vis, 0, sizeof(vis));
    memset(head, -1, sizeof(head));
    indx = 0;
    ip = 0;
}

void addedge(int u, int v, int w){
    edge[ip].v = v;
    edge[ip].w = w;
    edge[ip].next = head[u];
    head[u] = ip++;
}

void dfs(int u, int h){
    vis[u] = 1;
    ver[++indx] = u;
    deep[indx] = h;
    first[u] = indx;
    for(int i = head[u]; i != -1; i = edge[i].next){
        int v = edge[i].v;
        if(!vis[v]){
            dis[v] = dis[u] + edge[i].w;
            dfs(v, h + 1);
            ver[++indx] = u;
            deep[indx] = h;
        }
    }
}

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 x = dp[i][j - 1], y = dp[i + (1 << (j - 1))][j - 1];
            dp[i][j] = deep[x] < deep[y] ? x : y;
        }
    }
}

int RMQ(int l, int r){
    int len = log2(r - l + 1);
    int x = dp[l][len], y = dp[r - (1 << len) + 1][len];
    return deep[x] < deep[y] ? x : y;
}

int LCA(int x, int y){
    int l = first[x], r = first[y];
    if(l > r) swap(l, r);
    int pos = RMQ(l, r);
    return ver[pos];
}

int main(void){
    bool flag = false;
    int n, q, x, y, z;
    while(~scanf("%d", &n)){
        if(flag) puts("");
        flag = true;
        init();
        for(int i = 1; i < n; i++){
            scanf("%d%d%d", &x, &y, &z);
            addedge(x, y, z);
            addedge(y, x, z);
        }
        dis[1] = 0;
        dfs(1, 1);
        ST(2 * n - 1);
        scanf("%d", &q);
        while(q--){
            scanf("%d%d%d", &x, &y, &z);
            int lca1 = LCA(x, y);
            int lca2 = LCA(x, z);
            int lca3 = LCA(y, z);
            int sol1 = dis[x] + dis[y] - 2 * dis[lca1];
            int sol2 = dis[x] + dis[z] - 2 * dis[lca2];
            int sol3 = dis[y] + dis[z] - 2 * dis[lca3];
            printf("%d
", (sol1 + sol2 + sol3) >> 1);
        }
    }
    return 0;
}