제출 #266594

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
266594		Thistle	다리 (APIO19_bridges)	C++14	0 / 100	216 ms	61592 KiB

bridges

#pragma GCC target ("avx")
#pragma GCC optimize("Ofast")
#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_];
//--------------------------------------------------------------
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;
vec<P>edges;
bool chg[200000];
vec<P>qry;
int ans[200000];

vec<H>e[600000];

bool b[600000];

unionfind uf;


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(); }
    rep(i, siz(ed)) {
        e[edge[ed[i]].a].pb(H{ edge[ed[i]].b,ed[i] });
        e[edge[ed[i]].b].pb(H{ edge[ed[i]].a,ed[i] });
    }

    uf.init(n);
    int cur = 0;

    for (auto g : v) {
        //gは今回のクエリ　辿っていきます

        while (cur < siz(edges) && edge[edges[cur].zz].cost >= g.yy) {
            int k = edges[cur].zz;
            if (chg[k]) {
                cur++; continue;
            }

            int a = edge[k].a, b = edge[k].b;

            a = uf[a].fs, b = uf[b].fs;
            if (a == b) {
                cur++; 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

            for (auto g : e[a]) {
                e[b].pb(g);
            }

            cur++;
        }//グラフの縮約が完了した

        vec<int>vis;

        auto dfs = [&](int x, int cost, int num, auto dfs)->int {
            //今回はこれ以上でないといけない
            int sum = 0;
            b[x] = 1;
            vis.pb(x);
            sum += uf[x].sc;
            for (auto g : e[x]) {
                g.fs = uf[g.fs].fs;
                int ttt = (*prev(partition_point(all(edge[g.sc].v), [num](H a)->bool {return a.fs <= num; }))).sc;
                if (b[g.fs] || ttt < cost) 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
                sum += dfs(g.fs, cost, num, dfs);
            }
            return sum;
        };

        ans[g.zz] = dfs(uf[g.xx].fs, g.yy, g.zz, dfs);

        for (auto r : vis) b[r] = 0;
    }
}


signed main() {
    cin >> n >> m;


    rep(i, m) {
        int a, b, c;
        a = read(), b = read(), c = read();
        a--; b--;
        edge.pb(edg{ a,b,c,vec<H>(1,H{0,c}) });
        edges.pb(Q(a, b, i));
    }

    sort(all(edges), [&](P& a, P& b) {return edge[a.zz].cost > edge[b.zz].cost; });
    cin >> q;
    rep(i, q) {
        int t, a, b = 0;
        t = read(), a = read(), b = read();
        a--;
        qry.pb(Q(t, a, b));
        if (t == 1) {
            edge[a].v.pb(H{ i,b });
        }
    }

    
    int sq = sqrt(q) * 0.9;

    for (int i = 0; i < q; i += sq) {
        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 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;
            }
        }
        //sort(all(edges), [&](P& a, P& b) {return edge[a.zz].cost > edge[b.zz].cost; });
    }
    rep(i, q) {
        if (qry[i].xx == 2) printf("%lld\n", ans[i]);
    }
}

컴파일 시 표준 에러 (stderr) 메시지

bridges.cpp: In function 'll read()':
bridges.cpp:72:19: warning: unused variable 'k' [-Wunused-variable]
   72 | ll read() { ll u, k = scanf("%lld", &u); return u; }
      |                   ^
bridges.cpp: In function 'std::pair<long long int, long long int> readh(short int)':
bridges.cpp:74:33: warning: unused variable 'k' [-Wunused-variable]
   74 | 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; }
      |                                 ^

• Subtask 1: 0 / 13
• Subtask 2: 0 / 16
• Subtask 3: 0 / 17
• Subtask 4: 0 / 14
• Subtask 5: 0 / 13
• Subtask 6: 0 / 27