Codeforces Round #329 (Div. 2) D. Happy Tree Party-蒲公英云

Codeforces Round #329 (Div. 2) D. Happy Tree Party

Myth丶恋晨 2022-08-10 00:53 123阅读 0赞

对于从a→b 的路径，如果边权值为1，则值不变，否则就是1≤边权值，这样的话最多除64次就为0也就不需要继续除下去了，所以问题的关键在于去压缩哪些连续的边权为1的路径，这里自然就想到并查集啦，具体实现看代码吧！
另：因为更改后的边权值只会变小，所以只会涉及到集合并和查的操作。这一点要想懂。

// whn6325689
// Mr.Phoebe
// http://blog.csdn.net/u013007900
#include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <climits>
#include <complex>
#include <fstream>
#include <cassert>
#include <cstdio>
#include <bitset>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <ctime>
#include <set>
#include <map>
#include <cmath>
#include <functional>
#include <numeric>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
#define eps 1e-9
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define LLINF 1LL<<50
#define speed std::ios::sync_with_stdio(false);
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<ll, ll> pll;
typedef complex<ld> point;
typedef pair<int, int> pii;
typedef pair<pii, int> piii;
typedef vector<int> vi;
#define CLR(x,y) memset(x,y,sizeof(x))
#define CPY(x,y) memcpy(x,y,sizeof(x))
#define clr(a,x,size) memset(a,x,sizeof(a[0])*(size))
#define cpy(a,x,size) memcpy(a,x,sizeof(a[0])*(size))
#define debug(a) cout << #a" = " << (a) << endl;
#define debugarry(a, n) for (int i = 0; i < (n); i++) { cout << #a"[" << i << "] = " << (a)[i] << endl; }
#define mp(x,y) make_pair(x,y)
#define pb(x) push_back(x)
#define lowbit(x) (x&(-x))
#define MID(x,y) (x+((y-x)>>1))
#define getidx(l,r) (l+r | l!=r)
#define ls getidx(l,mid)
#define rs getidx(mid+1,r)
#define lson l,mid
#define rson mid+1,r
template<class T>
inline bool read(T &n)
{
    T x = 0, tmp = 1;
    char c = getchar();
    while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();
    if(c == EOF) return false;
    if(c == '-') c = getchar(), tmp = -1;
    while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar();
    n = x*tmp;
    return true;
}
template <class T>
inline void write(T n)
{
    if(n < 0)
    {
        putchar('-');
        n = -n;
    }
    int len = 0,data[20];
    while(n)
    {
        data[len++] = n%10;
        n /= 10;
    }
    if(!len) data[len++] = 0;
    while(len--) putchar(data[len]+48);
}
//-----------------------------------
const int MAXN = 10+2e5;
int n, m;
int dep[MAXN], st[MAXN];
ll w[MAXN];
vector<pii> g[MAXN];
pii pre[MAXN];
void dfs(int u,int fa)
{
    dep[u] = dep[fa]+1;
    for(auto it:g[u])
    {
        int v = it.first;
        int id = it.second;
        if(v == fa) continue;
        pre[v] = mp(u,id);
        dfs(v,u);
    }
}
int find_set(int u)
{
    if(w[pre[u].second] != 1) return u;
    if(st[u] == u)
        return st[u] = find_set(pre[u].first);
    return st[u] = find_set(st[u]);
}
int jmp(int a,ll &v)
{
    int id = pre[a].second;
    if(w[id] > 1)
    {
        v /= w[id];
        return pre[a].first;
    }
    return st[a] = find_set(st[a]);
}
ll cal(int a,int b,ll v)
{
    while(v && a!=b)
    {
        if(dep[a] < dep[b]) swap(a,b);
        a = jmp(a,v);
    }
    return v;
}
int main()
{
    // freopen("in.txt","r",stdin);
    scanf("%d%d",&n, &m);
    for(int i = 1; i < n; i ++)
    {
        int u,v;
        scanf("%d%d%I64d",&u,&v,&w[i]);
        g[u].pb(mp(v,i)), g[v].pb(mp(u,i));
    }
    dfs(1,0);
    for(int i = 1; i <= n; i ++) st[i] = i;
    while(m --)
    {
        int tp;
        scanf("%d",&tp);
        if(tp == 1)
        {
            int a, b;
            ll y;
            scanf("%d%d%I64d",&a, &b, &y);
            printf("%I64d\n",cal(a,b,y));
        }
        else
        {
            int p;
            ll c;
            scanf("%d%I64d",&p, &c);
            w[p] = c;
        }
    }
    return 0;
}