| # | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 | 
|---|---|---|---|---|---|---|---|
| 266855 | Thistle | 다리 (APIO19_bridges) | C++14 | 0 ms | 0 KiB | 
이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#pragma GCC target ("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#define _USE_MATH_DEFINES
#include<iostream>
#include<string>
#include<queue>
#include<cmath>
#include<map>
#include<set>
#include<list>
#include<iomanip>
#include<vector>
#include<random>
#include<functional>
#include<algorithm>
#include<stack>
#include<cstdio>
#include<cstring>
#include<bitset>
#include<unordered_set>
#include<unordered_map>
#include<climits>
#include<fstream>
#include<complex>
#include<time.h>
#include<cassert>
#include<functional>
#include<numeric>
#include<tuple>
using namespace std;
using ll = long long;
using ld = long double;
#define int long long
#define all(a) (a).begin(),(a).end()
#define fs first
#define sc second
#define xx first
#define yy second.first
#define zz second.second
#define H pair<int, int>
#define P pair<int, pair<int, int>>
#define Q(i,j,k) mkp(i,mkp(j,k))
#define rng(i,s,n) for(int i = (s) ; i < (n) ; i++)
#define rep(i,n) rng(i, 0, (n))
#define mkp make_pair
#define vec vector
#define vi vec<int>
#define pb emplace_back
#define siz(a) (int)(a).size()
#define crdcomp(b) sort(all((b)));(b).erase(unique(all((b))),(b).end())
#define getidx(b,i) (lower_bound(all(b),(i))-(b).begin())
#define ssp(i,n) (i==(int)(n)-1?"\n":" ")
#define ctoi(c) (int)(c-'0')
#define itoc(c) (char)(c+'0')
#define cyes printf("Yes\n")
#define cno printf("No\n")
#define cdf(n) int quetimes_=(n);rep(qq123_,quetimes_)
#define gcj printf("Case #%lld: ",qq123_+1)
#define readv(a,n) a.resize(n,0);rep(i,(n)) a[i]=read()
#define found(a,x) (a.find(x)!=a.end())
//#define endl "\n"
constexpr int mod = (ll)1e9 + 7;
constexpr int Mod = 998244353;
constexpr ld EPS = 1e-10;
constexpr ll inf = (ll)3 * 1e18;
constexpr int Inf = (ll)15 * 1e8;
constexpr int dx[] = { -1,1,0,0 }, dy[] = { 0,0,-1,1 };
template<class T>bool chmax(T& a, const T& b) { if (a < b) { a = b; return 1; } return 0; }
template<class T>bool chmin(T& a, const T& b) { if (b < a) { a = b; return 1; } return 0; }
ll read() { ll u, k = scanf("%lld", &u); return u; }
string reads() { string s; cin >> s; return s; }
H readh(short g = 0) { H u; int k = scanf("%lld %lld", &u.fs, &u.sc); if (g == 1) u.fs--, u.sc--; if (g == 2) u.fs--; return u; }
bool ina(H t, int h, int w) { return 0 <= t.fs && t.fs < h && 0 <= t.sc && t.sc < w; }
bool ina(int t, int l, int r) { return l <= t && t < r; }
ll gcd(ll i, ll j) { return j ? gcd(j, i % j) : i; }
ll popcount(ll x) {
    int sum = 0; for (int i = 0; i < 60; i++)if ((1ll << i) & x) sum++;
    return sum;
}
template<typename T>
class csum {
    vec<T> v;
public:
    csum(vec<T>& a) :v(a) { build(); }
    csum() {}
    void init(vec<T>& a) { v = a; build(); }
    void build() {
        for (int i = 1; i < v.size(); i++) v[i] += v[i - 1];
    }
    T a(int l, int r) {
        if (r < l) return 0;
        return v[r] - (l == 0 ? 0 : v[l - 1]);
    }//[l,r]
    T b(int l, int r) {
        return a(l, r - 1);
    }//[l,r)
    T a(pair<int, int>t) {
        return a(t.first, t.second);
    }
    T b(pair<int, int>t) {
        return b(t.first, t.second);
    }
};
class mint {
public:ll v;
      mint(ll v = 0) { s(v % mod + mod); }
      constexpr static int mod = (ll)1e9 + 7;
      constexpr static int fn_ = (ll)2e6 + 5;
      static mint fact[fn_], comp[fn_];
      mint pow(int x) const {
          mint b(v), c(1);
          while (x) {
              if (x & 1) c *= b;
              b *= b;
              x >>= 1;
          }
          return c;
      }
      inline mint& s(int vv) {
          v = vv < mod ? vv : vv - mod;
          return *this;
      }
      inline mint inv()const { return pow(mod - 2); }
      inline mint operator-()const { return mint() - *this; }
      inline mint& operator+=(const mint b) { return s(v + b.v); }
      inline mint& operator-=(const mint b) { return s(v + mod - b.v); }
      inline mint& operator*=(const mint b) { v = v * b.v % mod; return *this; }
      inline mint& operator/=(const mint b) { v = v * b.inv().v % mod; return *this; }
      inline mint operator+(const mint b) const { return mint(v) += b; }
      inline mint operator-(const mint b) const { return mint(v) -= b; }
      inline mint operator*(const mint b) const { return mint(v) *= b; }
      inline mint operator/(const mint b) const { return mint(v) /= b; }
      friend ostream& operator<<(ostream& os, const mint& m) {
          return os << m.v;
      }
      friend istream& operator>>(istream& is, mint& m) {
          int x; is >> x; m = mint(x);
          return is;
      }
      bool operator<(const mint& r)const { return v < r.v; }
      bool operator>(const mint& r)const { return v > r.v; }
      bool operator<=(const mint& r)const { return v <= r.v; }
      bool operator>=(const mint& r)const { return v >= r.v; }
      bool operator==(const mint& r)const { return v == r.v; }
      bool operator!=(const mint& r)const { return v != r.v; }
      explicit operator bool()const { return v; }
      explicit operator int()const { return v; }
      mint comb(mint k) {
          if (k > * this) return mint();
          if (!fact[0]) combinit();
          if (v >= fn_) {
              if (k > * this - k) k = *this - k;
              mint tmp(1);
              for (int i = v; i >= v - k.v + 1; i--) tmp *= mint(i);
              return tmp * comp[k.v];
          }
          return fact[v] * comp[k.v] * comp[v - k.v];
      }//nCk
      mint perm(mint k) {
          if (k > * this) return mint();
          if (!fact[0]) combinit();
          if (v >= fn_) {
              mint tmp(1);
              for (int i = v; i >= v - k.v + 1; i--) tmp *= mint(i);
              return tmp;
          }
          return fact[v] * comp[v - k.v];
      }//nPk
      static void combinit() {
          fact[0] = 1;
          for (int i = 1; i < fn_; i++) fact[i] = fact[i - 1] * mint(i);
          comp[fn_ - 1] = fact[fn_ - 1].inv();
          for (int i = fn_ - 2; i >= 0; i--) comp[i] = comp[i + 1] * mint(i + 1);
      }
}; mint mint::fact[fn_], mint::comp[fn_];
ll readi() {
    char c; bool st = 0, neg = 0;
    ll ret = 0;
    while (true) {
        c = getchar();
        if ((c - '0' < 0 || c - '0' > 9) && c != '-' && !st) continue;
        if ((c - '0' < 0 || c - '0' > 9) && c != '-' && st) break;
        if (st) ret *= 10;
        st = 1;
        if (c == '-') neg = 1;
        else ret += c - '0';
    }
    if (!neg) return ret;
    return -ret;
}
//--------------------------------------------------------------
class unionfind {
public:
    int size = 0;
    int pa[60000];
    void init(int n) {
        size = n;
        for (int i = 0; i <= size; i++) pa[i] = -1;
    }
    int find(int x) {
        if (pa[x] < 0) return x;
        return pa[x] = find(pa[x]);
    }
    bool unite(int x, int y) {
        x = find(x); y = find(y);
        if (x == y) return false;
        if (pa[x] > pa[y]) swap(x, y);
        pa[x] += pa[y];
        pa[y] = x;
        return true;
    }
    bool same(int x, int y) {
        return find(x) == find(y);
    }
    bool isroot(int x) {
        return x == find(x);
    }
    int sz(int x) {
        return -pa[find(x)];
    }
    H operator[](int x) {
        x = find(x);
        return H{ x,-pa[x] };
    }
};
//---------------------------------------------------------------------
int n, m, q;
struct edg {
    int a, b, cost;
    vec<H>v;//時刻、コスト
};
vec<edg>edge;
bool chg[200000];
int numb[60000];
vec<P>qry;
int ans[200000];
vec<H>e[600000];
bool b[600000];
unionfind uf;
vi idx;
vi numbers[300000];
void solve(vec<P>& v, vi& ed) {
    sort(all(v), [](P& a, P& b) {return a.yy > b.yy; });
    //大きさが大きい順にソートをした
    rep(i, n) { e[i].clear(); numb[i] = i; }
    for (auto& g : ed) {
        e[edge[g].a].pb(H{ edge[g].b,g });
        e[edge[g].b].pb(H{ edge[g].a,g });
    }
    uf.init(n);
    int cur = siz(idx) - 1;
    for (auto& g : v) {
        //gは今回のクエリ 辿っていきます
        while (cur >= 0 && idx[cur] >= g.yy) {
            for (auto& rt : numbers[cur]) {
                int k = rt;
                if (chg[k]) continue;
                int a = edge[k].a, b = edge[k].b;
                a = uf[a].fs, b = uf[b].fs;
                if (a == b) continue;
                uf.unite(b, a);
                if (uf[a].fs == a) swap(a, b);
                //aが親を変えられたうらみのある人
                if (siz(e[numb[b]]) > siz(e[numb[a]])) {
                    for (auto& g : e[numb[a]]) {
                        e[numb[b]].pb(g);
                    }
                }
                else {
                    for (auto& g : e[numb[b]]) {
                        e[numb[a]].pb(g);
                    }
                    numb[b] = numb[a];
                }
            }
            cur--;
        }//グラフの縮約が完了した
        vi vis;
        queue<int>q;
        q.push(g.fs);
        int sum = 0;
        //今回はこれ以上でないといけない
        while (!q.empty()) {
            int x = q.front(); q.pop();
            if (b[x]) continue;
            b[x] = 1;
            vis.pb(x);
            sum += uf[x].sc;
            for (auto& g : e[numb[x]]) {
                g.fs = uf[g.fs].fs;
                if (b[g.fs] || (*prev(partition_point(all(edge[g.sc].v), [g.zz](H a)->bool {return a.fs <= g.zz; }))).sc < g.yy) continue;
                //訪れたことがない、二分探索をした時のコストが自分よりも少なくない
                q.push(g.fs);
            }
        }
        ans[g.zz] = sum;
        for (auto& r : vis) b[r] = 0;
    }
}
signed main() {
    n = readi(), m = readi();
    int t, a, b, c;
    edge.resize(m);
    rep(i, m) {
        a = readi(), b = readi(), c = readi();
        a--; b--;
        idx.pb(c);
        edge[i] = edg{ a,b,c,vec<H>(1,H{0,c}) };
    }
    q = readi();
    qry.resize(q);
    rep(i, q) {
        t = readi(), a = readi(), b = readi();
        a--;
        qry[i] = Q(t, a, b);
        if (t == 1) {
            edge[a].v.pb(H{ i,b });
            idx.pb(b);
        }
    }
    crdcomp(idx);
    rep(i, m) {
        numbers[getidx(idx, edge[i].cost)].pb(i);
    }
    int sq = sqrt(q) * 2;
    vec<P>qq; //どっちのクエリか、重量がどうなっているのか、次の情報
    vi p;//道
    for (int i = 0; i < q; i += sq) {
        qq.clear(); p.clear();
        for (int j = i; j < i + sq && j < q; j++) {
            if (qry[j].xx == 1) {
                chg[qry[j].yy] = 1;
                p.pb(qry[j].yy);
            }
            else {
                qq.pb(Q(qry[j].yy, qry[j].zz, j));
            }
        }
        solve(qq, p);
        for (int j = i; j < i + sq && j < q; j++) {
            if (qry[j].xx == 1) {
                chg[qry[j].yy] = 0;
                edge[qry[j].yy].cost = qry[j].zz;
                numbers[getidx(idx, qry[j].zz)].pb(qry[j].yy);
            }
        }
        rep(j, siz(idx)) {
            int sum = 0;
            for (auto& k : numbers[j]) {
                if (edge[k].cost != idx[j]) {
                    sum++;
                }
            }
            if (sum) {
                vi v(siz(numbers[j]) - sum);
                sum = 0;
                for (auto& k : numbers[j]) {
                    if (edge[k].cost == idx[j]) {
                        v[sum++] = k;
                    }
                }
                numbers[j] = v;
            }
        }
    }
    rep(i, q) {
        if (qry[i].xx == 2) printf("%lld\n", ans[i]);
    }
}
