CDOJ 92 Journey(LCA&RMQ)

题目连接：http://acm.uestc.edu.cn/#/problem/show/92

题意：给定一棵树，最后给加一条边，给定Q次查询，每次查询加上最后一条边之后是否比不加这条边要近，如果近的话，输出近多少，否则输出0

思路：没加最后一条边之前两点之间的距离是dis(u) + dis(v) - 2*dis(lca(u, v)); 加上之后就是必须要经过这两个点。(假设最后添加的一条边的端点为x和y，权值为w)min(dis(u, x) + dis(v, y) + w, dis(u, y) + dis(v, x) + w)

代码如下：

/*************************************************************************
    > File Name:            4.cpp
    > Author:               Howe_Young
    > Mail:                 1013410795@qq.com
    > Created Time:         2015年10月08日 星期四 19时53分38秒
 ************************************************************************/

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>

using namespace std;
typedef long long ll;
const int maxn = 1e5 + 20;
struct Edge {
    int to, next, w;
}edge[maxn<<1];
int tot, head[maxn];
int cnt;
int Euler[maxn<<1];
int R[maxn];
int dis[maxn<<1];
int dep[maxn<<1];
int dp[maxn<<1][20];
void init()
{
    cnt = 0;
    tot = 0;
    memset(head, -1, sizeof(head));
}
void addedge(int u, int v, int w)
{
    edge[tot].to = v;
    edge[tot].w = w;
    edge[tot].next = head[u];
    head[u] = tot++;
}
void dfs(int u, int fa, int depth, int dist)
{
    Euler[++cnt] = u;
    R[u] = cnt;
    dep[cnt] = depth;
    dis[cnt] = dist;
    for (int i = head[u]; i != -1; i = edge[i].next)
    {
        int v = edge[i].to;
        if (v == fa) continue;
        dfs(v, u, depth + 1, dist + edge[i].w);
        Euler[++cnt] = u;
        dep[cnt] = depth;
        dis[cnt] = dist;
    }
}

void RMQ(int n)
{
    for (int i = 1; i <= n; i++) dp[i][0] = i;
    int m = log2(n);
    for (int j = 1; j <= m; j++)
        for (int i = 1; i + (1 << j) - 1 <= n; i++)
            dp[i][j] = dep[dp[i][j - 1]] < dep[dp[i + (1 << (j - 1))][j - 1]] ? dp[i][j - 1] : dp[i + (1 << (j - 1))][j - 1];
}
int getLCA(int u, int v)
{
    if (u == v) return v;
    int l = R[u], r = R[v];
    if (l > r) swap(l, r);
    int k = log2(r - l + 1);
    int lca = dep[dp[l][k]] < dep[dp[r - (1 << k) + 1][k]] ? dp[l][k] : dp[r - (1 << k) + 1][k];
    return Euler[lca];
}
int getdist(int u, int v)
{
    if (u == v) return 0;
    int l = R[u], r = R[v];
    int lca = getLCA(u, v);
    return dis[l] + dis[r] - 2 * dis[R[lca]];
}
int main()
{
    int T, kase = 0;
    scanf("%d", &T);
    while (T--)
    {
        init();
        int n, Q;
        int u, v, w;
        scanf("%d %d", &n, &Q);
        for (int i = 1; i < n; i++)
        {
            scanf("%d %d %d", &u, &v, &w);
            addedge(u, v, w);
            addedge(v, u, w);
        }
        scanf("%d %d %d", &u, &v, &w);
        dfs(1, 0, 1, 0);
        RMQ(cnt);
        printf("Case #%d:
", ++kase);
        while (Q--)
        {
            int a, b;
            scanf("%d %d", &a, &b);
            int dis1 = getdist(a, b);
            int dis2 = min(getdist(a, u) + getdist(v, b) + w, getdist(a, v) + getdist(u, b) + w);
            if (dis1 > dis2)
                printf("%d
", dis1 - dis2);
            else
                printf("0
");
        }
    }
    return 0;
}