Uva - 11354 Bond 【多次询问最小瓶颈路 and LCA预处理】

传送门
题意: 跟定一副加权无向图, q次询问, 每次询问两个点之间的路径中最大边的最小是多少.

思路: 很明显, 这个是求最小瓶颈路, 只不过是多次询问, 那么我们可以知道答案一定实在MST上, 所以先求MST, 然后用LCA预处理这颗树, 就可以做到logn的回答了.

复杂度: O(mlogm + nlogn + qlogn)
AC Code

const int maxn = 1e5+5;
// LCA
int up[maxn][23], maxx[maxn][23];
int deep[maxn], dis[maxn];
int cnt, head[maxn];
int n, m, q;
struct node {
    int to, next, w;
}e[maxn<<1];
void init() {
    Fill(head,-1); Fill(dis,0);
    Fill(up,0);  Fill(deep,0);
    cnt = 0;  Fill(maxx, 0);
}
void add(int u, int v, int w) {
    e[cnt] = node{v, head[u], w};
    head[u] = cnt++;
}

void dfs(int u,int fa,int d)
{
    deep[u] = d + 1;
    for(int i = 1 ; i < 20 ; i ++) {
        up[u][i] = up[up[u][i-1]][i-1];
        maxx[u][i] = max(maxx[up[u][i-1]][i-1], maxx[u][i-1]);
    }
    for(int i = head[u] ; ~i ; i = e[i].next) {
        int to = e[i].to;
        if(to == fa) continue;
        dis[to] = dis[u] + e[i].w;
        up[to][0] = u;
        maxx[to][0] = e[i].w;
        dfs(to, u, d+1);
    }
}

int LCA_BZ(int u,int v)
{
    int mx = 0;
    if(deep[u] < deep[v]) swap(u,v);
    int k = deep[u] - deep[v];
    for(int i = 0 ; i < 20 ; i ++) {
        if((1<if(u != v) {
        for(int i = 19 ; i >= 0 ; i --) {
            if(up[u][i] != up[v][i]) {
                mx = max(mx, max(maxx[v][i], maxx[u][i]));
                u = up[u][i];
                v = up[v][i];
            }
        }
        mx = max(mx, max(maxx[u][0], maxx[v][0]));
        u = up[u][0];
    }
    return mx;
}

// MST
struct edge {
    int  u, v, w;
    bool operator < (const edge &_ ) const {
        return w < _.w;
    }
}s[maxn];
int fa[maxn], r[maxn];
bool vis[maxn];
void init2() {
    for (int i = 1 ; i <= n ; i ++) {
        fa[i] = i;
        r[i] = 1;
    }
    Fill(vis, 0);
}
int Find(int x) {
    return fa[x] == x ? x : fa[x] = Find(fa[x]);
}
bool Un(int x, int y) {
    int fx = Find(x);
    int fy = Find(y);
    if (fx == fy) return false;
    if (r[fx] > r[fy]) swap(fx, fy);
    fa[fx] = fy;
    r[fy] += r[fx];
    return true;
}

void kru() {
    int cnt = 0;
    for (int i = 1 ; i <= m ; i ++) {
        if (Un(s[i].u, s[i].v)) {
            ++ cnt;
            add(s[i].u, s[i].v, s[i].w);
            add(s[i].v, s[i].u, s[i].w);
            vis[i] = 1;
        }
        if (cnt >= n - 1) break;
    }
}
void solve()
{
    int f = 0;
    while(~scanf("%d%d",&n,&m)) {
        if (f) printf("\n"); f = 1;
        init(); init2();
        for (int i = 1 ; i <= m ; i ++) {
            scanf("%d%d%d",&s[i].u, &s[i].v, &s[i].w);
        }
        sort(s+1, s+1+m);
        kru(); up[1][0] = 1;   //那根节点的父亲设为他自己.
        maxx[1][0] = inf; dfs(1,-1,0);
        scanf("%d", &q);
        while(q--) {
            int s, t;
            scanf("%d%d", &s, &t);
            printf("%d\n", LCA_BZ(s, t));
        }
    }
}