UESTC 92 Journey （LCA）

题意:

给出N<=105的一棵树, 和一条新建的边, Q<=105询问两点间节省了多少时间

分析:

知道dis[u][v]=dis[u]+dis[v]−2∗dis[lca(u,v)]就好做辣

代码:

////  Created by TaoSama on 2015-09-24//  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, q, a, b, l;int cnt, head[N], pnt[N << 1], nxt[N << 1], cost[N << 1];void add_edge(int u, int v, int c) {    pnt[cnt] = v;    cost[cnt] = c;    nxt[cnt] = head[u];    head[u] = cnt++;}int dp[N], p[18][N], dep[N];void dfs(int u, int f, int c, int d) {    p[0][u] = f;    dp[u] = c;    dep[u] = d;    for(int i = head[u]; ~i; i = nxt[i]) {        int v = pnt[i];        if(v == f) continue;        dfs(v, u, c + cost[i], d + 1);    }}void gao() {    dfs(1, -1, 0, 0);    for(int i = 0; i + 1 < 18; ++i)        for(int v = 1; v <= n; ++v)            if(p[i][v] < 0) p[i + 1][v] = -1;            else p[i + 1][v] = p[i][p[i][v]];}int lca(int u, int v) {    if(dep[u] > dep[v]) swap(u, v);    for(int i = 0; i < 18; ++i)        if(dep[u] - dep[v] >> i & 1)            v = p[i][v];    if(u == v) return u;    for(int i = 18 - 1; ~i; --i) {        if(p[i][u] != p[i][v]) {            u = p[i][u];            v = p[i][v];        }    }    return p[0][u];}int main() {#ifdef LOCAL    freopen("in.txt", "r", stdin);//  freopen("out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    int t; scanf("%d", &t);    int kase = 0;    while(t--) {        scanf("%d%d", &n, &q);        cnt = 0; memset(head, -1, sizeof head);        for(int i = 1; i < n; ++i) {            int u, v, c; scanf("%d%d%d", &u, &v, &c);            add_edge(u, v, c);            add_edge(v, u, c);        }        scanf("%d%d%d", &a, &b, &l);        printf("Case #%d:\n", ++kase);        gao();        while(q--) {            int u, v; scanf("%d%d", &u, &v);            int uv = dp[u] + dp[v] - 2 * dp[lca(u, v)];            int ua = dp[u] + dp[a] - 2 * dp[lca(u, a)];            int bv = dp[b] + dp[v] - 2 * dp[lca(b, v)];            int ub = dp[u] + dp[b] - 2 * dp[lca(u, b)];            int av = dp[a] + dp[v] - 2 * dp[lca(a, v)];            int ans = max(0, uv - min(ua + bv, ub + av) - l);            printf("%d\n", ans);        }    }    return 0;}