POJ 1986Distance Queries（LCA）

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题意:

给定的一棵树,边数卖萌的,(u,v,c,d),d方向也是卖萌的
K≤104次询问,dis(u,v)

分析:

树上2点之间距离dis(u,v)=dis(u)+dis(v)−2∗dis(lca(u,v))
剩下就是lca模版题

代码:

////  Created by TaoSama on 2015-11-09//  Copyright (c) 2015 TaoSama. All rights reserved.////#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << "  "#define prln(x) cout << #x << " = " << x << endlconst int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;int n, m, q;int vs[N << 1], dep[N << 1], first[N], dfsNum;int getMin(int x, int y) {    return dep[x] < dep[y] ? x : y;}struct SparseTable {    int dp[20][N << 1];    void init(int n) {        for(int i = 1; i <= n; ++i) dp[0][i] = i;        for(int i = 1; 1 << i <= n; ++i)            for(int j = 1; j + (1 << i) - 1 <= n; ++j)                dp[i][j] = getMin(dp[i - 1][j], dp[i - 1][j + (1 << i - 1)]);    }    int RMQ(int l, int r) {        int k = 31 - __builtin_clz(r - l + 1);        return getMin(dp[k][l], dp[k][r - (1 << k) + 1]);    }} st;struct Edge {    int v, c;};vector<Edge> G[N];int dp[N];void dfs(int u, int f, int d) {    vs[++dfsNum] = u;    dep[dfsNum] = d;    first[u] = dfsNum;    for(int i = 0; i < G[u].size(); ++i) {        Edge &e = G[u][i];        if(e.v == f) continue;        dp[e.v] = dp[u] + e.c;        dfs(e.v, u, d + 1);        vs[++dfsNum] = u;        dep[dfsNum] = d;    }}void init() {    dfsNum = 0;    dfs(1, -1, 0);    st.init(2 * n - 1);}int lca(int u, int v) {    if(first[u] > first[v]) swap(u, v);    int idx = st.RMQ(first[u], first[v]);    return vs[idx];}int main() {#ifdef LOCAL    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    while(scanf("%d%d", &n, &m) == 2) {        for(int i = 1; i <= n; ++i) G[i].clear();        while(m--) {            int u, v, c, d[2]; scanf("%d%d%d%s", &u, &v, &c, d);            G[u].push_back((Edge) {v, c});            G[v].push_back((Edge) {u, c});        }        init();        scanf("%d", &q);        while(q--) {            int u, v; scanf("%d%d", &u, &v);            int _lca = lca(u, v);            printf("%d\n", dp[u] + dp[v] - 2 * dp[_lca]);        }    }    return 0;}

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