牛客小白月赛13 F题

链接：https://ac.nowcoder.com/acm/contest/549/F
来源：牛客网

小A这次来到一个景区去旅游，景区里面有N个景点，景点之间有N-1条路径。小A从当前的一个景点移动到下一个景点需要消耗一点的体力值。但是景区里面有两个景点比较特殊，它们之间是可以直接坐观光缆车通过，不需要消耗体力值。而小A不想走太多的路，所以他希望你能够告诉它，从当前的位置出发到他想要去的那个地方，他最少要消耗的体力值是多少。

思路：LCA（不懂得百度）裸题

作者：西格玛象限
链接：https://ac.nowcoder.com/discuss/177449?type=101&order=0&pos=7&page=1
来源：牛客网

这张图可以认为是边权全为1的树上增加了一条边权为0的边。

首先先不考虑多出来的一条边，那么dep[u]表示点u的深度，任意两点u，v的最短的距离就是
dep[u]+dep[v]−2dep[lca(u,v)],
加上这条边后另外一种可能最短的路径就是经过这条边权为0的边，可以建图之后跑一次最短路，每次查询的答案都是
min(dis[u]+dis[v],dep[u]+dep[v]−2dep[lca(u,v)])。

#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e6 + 5;
int head[maxn], tot;
struct Edge {
    int u, v, next;
}edge[maxn];
int father[maxn], depth[maxn], size[maxn], son[maxn], top[maxn];
int L[maxn], R[maxn], Index;
void init() {
    memset(head, -1, sizeof(head));
    tot = 1;
}
 
void add(int u, int v) {
    edge[++tot].u = u; edge[tot].v = v;
    edge[tot].next = head[u];
    head[u] = tot;
}
 
void dfs1(int u, int fa) {
    size[u] = 1;
    son[u] = 0;
    father[u] = fa;
    depth[u] = depth[fa] + 1;
    int maxson = -1;
    for (int i = head[u]; i != -1; i = edge[i].next) {
        int to = edge[i].v;
        if (to == fa) {
            continue;
        }
        dfs1(to, u);
        size[u] += size[to];
        if (size[to] > maxson) {
            maxson = size[to];
            son[u] = to;
        }
    }
}
 
void dfs2(int u, int topf) {
    top[u] = topf;
    L[u] = R[u] = ++Index;
    if (son[u]) {
        dfs2(son[u], topf);
    }
    for (int i = head[u]; i != -1; i = edge[i].next) {
        int to = edge[i].v;
        if (L[to] || to == son[u]) {
            continue;
        }
        dfs2(to, to);
    }
    R[u] = Index;
}
 
int LCA(int x, int y) {
    while (top[x] != top[y]) {
        if (depth[top[x]] < depth[top[y]]) {
            swap(x, y);
        }
        x = father[top[x]];
    }
    if (depth[x] > depth[y]) {
        return y;
    }
    return x;
}
 
 
 
int main() {
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
    init();
    int n, m, s, u, v;
    scanf("%d", &n);
    for (int i = 1; i <= n - 1; i++) {
        scanf("%d%d", &u, &v);
        add(u, v); add(v, u);
    }
    dfs1(1, 1);
    dfs2(1, 1);
    int a, b, c;
	scanf("%d%d", &a, &b);
	int x, y;
    scanf("%d", &m);
    for (int i = 1; i <= m; i++) {
        scanf("%d%d", &x, &y);
        int L1 = LCA(x, y);
        int L2 = LCA(x, a);
        int L3 = LCA(x, b);
        int L4 = LCA(y, a);
        int L5 = LCA(y, b);
//        cout << L1 << " " << L2 << " " << L3 << " " << L4 << " " << L5 << endl;
        int x1 = depth[x] + depth[y] - 2 * depth[L1];
        int x2 = depth[x] + depth[a] - 2 * depth[L2];
        int x3 = depth[y] + depth[b] - 2 * depth[L5];
        int x4 = depth[x] + depth[b] - 2 * depth[L3];
        int x5 = depth[y] + depth[a] - 2 * depth[L4];
//        cout << x1 << " " << x2 << " " << x3 << " " << x4 << " " << x5 << endl;
        int tmp = min(x2 + x3, x4 + x5);
        printf("%d\n", min(tmp, x1));
         
    }
    return 0;
}