HDU 2887 Watering Hole(MST + 倍增LCA)

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总算是做上一道LCA的应用题了...

题意：有$n$个牧场， $m$根管道分别连接编号为$u,v$的牧场花费$p_{i}$，在第$i$个牧场挖口井需要花费$w_{i}$，有$P$根管道直接连通着$u,v$，即免费连上$u,v$

对每根免费管道输出让所有牧场都有水的最小花费

先是最小生成树，用0去连每一个点，边权就是每个点的权值，然后正常建，跑一遍最小生成树，把用到的边重新建一次图，然后就对每次询问的$u,v$，减掉他们之间的路径的最长边就是答案了

因为删去这其中一条边，再免费连上$u,v$，最后还是一棵树，最小花费就减去最长边就好了。

然后求两点路径上的最长边就得用到倍增LCA，本来有点想不太明白，然后画了个图就清楚了，再预处理的dfs中，求每个点的LCA就可以直接求最长边了

$cost[u][i]$表示u往上走$2^{i}$步之间的最长边

转移就是$cost[u][i]=\max (cost[u][i-1], cost[lca[u][i-1]][i-1])$
求出来的是$u$往上走1, 2, 4, ...步的距离

如果需要往上走3步的答案即为$max(cost[u][0], cost[lca[u][0]][1])$

这在查询过程中实现即可

#include <bits/stdc++.h>

using namespace std;

inline int read() {

    int x = , f = ; char ch = getchar();

    while (ch < '' || ch > '') { if (ch == '-') f = -; ch = getchar(); }

    while (ch >= '' && ch <= '') { x = x *  + ch - ; ch = getchar(); }

    return x * f;

}

const int N = 5e3 + ;

const int M = 3e5 + ;

struct Edge1 {

    int u, v, c;

    bool operator < (const Edge1 &rhs) const {

        return c < rhs.c;

    }

} e[N + M];

struct Edge {int v, next, c;} edge[*N];

int cnt, head[N], fa[N], lca[N][], cost[N][], dep[N], n, m, q;

bool vis[N];

inline void addedge(int u, int v, int c) {

    edge[cnt].v = v; edge[cnt].c = c; edge[cnt].next = head[u]; head[u] = cnt++;

}

void init() {

    cnt = ;

    for (int i = ; i <= n; i++) fa[i] = i, head[i] = -, dep[i] = , vis[i] = false;

    memset(cost, , sizeof(cost));

    memset(lca, , sizeof(lca));

}

int getfa(int x) { return x == fa[x] ? x : fa[x] = getfa(fa[x]); }

void dfs(int u) {

    lca[u][] = fa[u];

    vis[u] = ;

    for (int i = ; i <= ; i++) {

        lca[u][i] = lca[lca[u][i-]][i-];

        cost[u][i] = max(cost[u][i-], cost[lca[u][i-]][i-]);

    }

    for (int i = head[u]; ~i; i = edge[i].next) {

        int v = edge[i].v, c = edge[i].c;

        if (vis[v]) continue;

        dep[v] = dep[u] + ;

        cost[v][] = c;

        fa[v] = u;

        dfs(v);

    }

}

int Lca(int u, int v) {

    if (dep[u] < dep[v]) swap(u, v);

    int ans = ;

    int f = dep[u] - dep[v];

    for (int i = ; i <= ; i++) {

        if (f & ( << i)) {

            ans = max(ans, cost[u][i]);

            u = lca[u][i];

        }

    }

    if (u == v) return ans;

    for (int i = ; i >= ; i--) {

        if (lca[u][i] != lca[v][i]) {

            ans = max(ans, cost[u][i]);

            ans = max(ans, cost[v][i]);

            u = lca[u][i];

            v = lca[v][i];

        }

    }

    return max(max(ans, cost[u][]), cost[v][]);

}

int main() {

    while (~scanf("%d%d%d", &n, &m, &q)) {

        init();

        int sum = ;

        int tol = ;

        for (int i = ; i <= n; i++) {

            int p = read();

            e[++tol].u = ; e[tol].v = i; e[tol].c = p;

        }

        for (int i = ; i <= m; i++) {

            int u = read(), v = read(), c = read();

            e[++tol].u = u, e[tol].v = v, e[tol].c = c;

        }

        sort(e + , e +  + tol);

        for (int i = ; i <= tol; i++) {

            int u = getfa(e[i].u), v = getfa(e[i].v);

            if (u == v) continue;

            fa[v] = u;

            sum += e[i].c;

            addedge(e[i].u, e[i].v, e[i].c);

            addedge(e[i].v, e[i].u, e[i].c);

        }

        fa[] = ;

        dfs();

        while (q--) {

            int u = read(), v = read();

            printf("%d\n", sum - Lca(u, v));

        }

    }

    return ;

}

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HDU 2887 Watering Hole(MST + 倍增LCA)

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