SPOJ QTREE Query on a tree (树链剖分+线段树)

题意：给定一棵树，然后有一些操作，有两种，一种是改变某条边的权值，第二种是询问从u->v的路径中，边权最大的是多少。

析：就一个树链剖分，然后用线段树维护即可。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e16;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 20000 + 10;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}

struct Edge{
  int to, next;
};
Edge edges[maxn<<1];
int p[maxn], fp[maxn], son[maxn], top[maxn];
int dp[maxn], fa[maxn], head[maxn], num[maxn];
int pos, cnt;
int maxv[maxn<<2];

void init(){
  pos = 1;  cnt = 0;
  memset(maxv, 0, sizeof maxv);
  memset(son, -1, sizeof son);
  memset(head, -1, sizeof head);
}

void add_edge(int u, int v){
  edges[cnt].to = v;
  edges[cnt].next = head[u];
  head[u] = cnt++;
}

void dfs1(int u, int pre, int d){
  dp[u] = d;  fa[u] = pre;
  num[u] = 1;
  for(int i = head[u]; ~i; i = edges[i].next){
    int v = edges[i].to;
    if(v == pre)  continue;
    dfs1(v, u, d+1);
    num[u] += num[v];
    if(son[u] == -1 || num[v] > num[son[u]])  son[u] = v;
  }
}

void dfs2(int u, int sp){
  top[u] = sp;  p[u] = pos++;
  fp[p[u]] = u;
  if(son[u] == -1)  return ;
  dfs2(son[u], sp);
  for(int i = head[u]; ~i; i = edges[i].next){
    int v = edges[i].to;
    if(v != fa[u] && v != son[u])  dfs2(v, v);
  }
}
int e[maxn][3];

void update(int x, int val, int l, int r, int rt){
  if(x == l && r == x){
    maxv[rt] = val;
    return ;
  }
  int m = l+r >> 1;
  if(x <= m)  update(x, val, lson);
  else  update(x, val, rson);
  maxv[rt] = max(maxv[rt<<1], maxv[rt<<1|1]);
}

int query(int L, int R, int l, int r, int rt){
  if(L <= l && r <= R)  return maxv[rt];
  int ans = 0;
  int m = l+r >> 1;
  if(L <= m)  ans = max(ans, query(L, R, lson));
  if(R > m)   ans = max(ans, query(L, R, rson));
  return ans;
}

int solve(int u, int v){
  int ans = 0;
  int f1 = top[u], f2 = top[v];
  while(f1 != f2){
    if(dp[f1] < dp[f2]){
      swap(f1, f2);  swap(u, v);
    }
    ans = max(ans, query(p[f1], p[u], 1, n, 1));
    u = fa[f1];
    f1 = top[u];
  }

  if(u == v)  return ans;
  if(dp[u] > dp[v])  swap(u, v);
  return max(ans, query(p[son[u]], p[v], 1, n, 1));
}


int main(){
  int T;  cin >> T;
  while(T--){
    scanf("%d", &n);
    init();
    for(int i = 1; i < n; ++i){
      scanf("%d %d %d", &e[i][0], &e[i][1], &e[i][2]);
      add_edge(e[i][0], e[i][1]);
      add_edge(e[i][1], e[i][0]);
    }
    dfs1(1, 1, 0);
    dfs2(1, 1);
    for(int i = 1; i < n; ++i){
      if(dp[e[i][0]] > dp[e[i][1]])  swap(e[i][0], e[i][1]);
      update(p[e[i][1]], e[i][2], 1, n, 1);
    }
    char s[10];
    while(scanf("%s", s) == 1 && s[0] != 'D'){
      int u, v;
      scanf("%d %d", &u, &v);
      if(s[0] == 'Q')  printf("%d\n", solve(u, v));
      else  update(p[e[u][1]], v, 1, n, 1);
    }
  }
  return 0;
}