poj2763 Housewife Wind--LCA转RMQ

原题链接：​​http://poj.org/problem?id=2763​​

题意：n个点，n-1条点的连线，数据保证任意两点可达，无环，接下来q行操作，两种形式，0 u 表示查询该人到u的时间；1 i w 表示第i条路的时间改为w。

分析：只能用RMQ来做，有点细节处理，都在注释里。

#define _CRT_SECURE_NO_DEPRECATE 

#include<iostream>
#include<vector>
#include<cstring>
#include<queue>
#include<stack>
#include<algorithm>
#include<cmath>
#define INF 99999999
#define eps 0.0001
#define N ((1<<12)+10)
using namespace std;

struct Edge
{
  int v;
  int index;
  int next;
};

int n, q, s;
int index;
int cnt;
Edge edge[2 * 100005];
int head[100005];
int dis[100005];
int father[100005];//父节点
int time[100005];//第i条路的时间
int node[2 * 100005];//保存第i次访问的节点
int first[100005];//i这个节点是第几次访问到的
int depth[2 * 100005];//第i次访问的节点的深度
int dp[2 * 100005][25];//dp[i][j]表示从第i次访问开始，连续2^j个访问内，哪次访问的节点深度最小
bool vis[100005];


void add(int u, int v, int index)
{
  edge[cnt].v = v; edge[cnt].index = index; edge[cnt].next = head[u]; head[u] = cnt++;
  edge[cnt].v = u; edge[cnt].index = index; edge[cnt].next = head[v]; head[v] = cnt++;
}

void dfs(int u, int dep, int fa)
{
  index++;
  vis[u] = 1;
  first[u] = index;
  node[index] = u;
  depth[index] = dep;
  dis[u] = dep;
  father[u] = fa;

  for (int i = head[u]; i !=-1; i=edge[i].next)
  {
    int v = edge[i].v;
    if (!vis[v])
    {
      dfs(v, dep + time[edge[i].index], u);
      index++;
      node[index] = u;
      depth[index] = dep;
    }
  }
}

void ST(int n)
{
  int k = log((double)n) / log(2.0);
  for (int i = 1; i <= n; i++)
    dp[i][0] = i;

  for (int j = 1; j <= k; j++)
    for (int i = 1; i + (1 << j) - 1 <= n; i++)
    {
      int a = dp[i][j - 1];
      int b = dp[i + (1 << (j - 1))][j - 1];
      if (depth[a] < depth[b])
        dp[i][j] = a;
      else
        dp[i][j] = b;
    }
}

int LCA(int u, int v)
{
  int i = first[u];
  int j = first[v];
  if (i > j)
    swap(i, j);

  int k = log(j - i + 1.0) / log(2.0);
  int a = dp[i][k];
  int b = dp[j - (1 << k) + 1][k];
  int x = (depth[a] < depth[b]) ? a : b;
  return node[x];
}

void solve(int u,int v)
{
  int lca = LCA(u, v);
  printf("%d\n", dis[u] + dis[v] - 2 * dis[lca]);
}

void update(int u,int t,int fa)
{
  for (int i = head[u]; i != -1; i = edge[i].next)
  {
    int v = edge[i].v;
    if (father[v] == fa)
    {
      dis[v] += t;
      update(v, t, v);
    }
  }
}

int main()
{
  int u, v;
  int op;
  while (~scanf("%d%d%d", &n, &q, &s))
  {
    index = cnt = 0;
    memset(head, -1, sizeof(head));
    memset(vis, 0, sizeof(vis));
    depth[1] = 0;
    dis[1] = 0;
    memset(dp, 0, sizeof(dp));

    for (int i = 1; i < n; i++)
    {
      scanf("%d%d%d", &u, &v, &time[i]);
      add(u, v, i);
    }

    dfs(1, 0, 1);
    ST(index);

    while (q--)
    {
      scanf("%d", &op);
      if (op)
      {
        int ith, w;
        scanf("%d%d", &ith, &w);

        int pos = (ith - 1) * 2;
        int x = edge[pos].v;
        int y = edge[pos + 1].v;

        int ww = time[ith];
        time[ith] = w;//数据要更新，记住

        if (dis[x] > dis[y])
          swap(x, y);

        int t = w - ww;
        dis[y] += t;
        update(y, t, y);
      }
      else
      {
        int x;
        scanf("%d", &x);
        solve(s, x);
        s = x;
      }
    }
  }
  return 0;
}