結果

問題 No.1502 Many Simple Additions
ユーザー au7777
提出日時 2022-05-05 01:06:17
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
WA  
実行時間 -
コード長 4,062 bytes
コンパイル時間 4,279 ms
コンパイル使用メモリ 245,776 KB
実行使用メモリ 17,872 KB
最終ジャッジ日時 2024-07-04 03:33:01
合計ジャッジ時間 7,155 ms
ジャッジサーバーID
（参考情報） 		judge2 / judge3
このコードへのチャレンジ
（要ログイン）
ファイルパターン 結果
sample AC * 5
other AC * 14 WA * 25
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

#include <bits/stdc++.h>
#include <atcoder/all>
typedef long long int ll;
typedef long long int ull;
#define MP make_pair
using namespace std;
using namespace atcoder;
typedef pair<ll, ll> P;
// const ll MOD = 998244353;
const ll MOD = 1000000007;
// using mint = modint998244353;
using mint = modint1000000007;
const double pi = 3.1415926536;
const int MAX = 2000005;
long long fac[MAX], finv[MAX], inv[MAX];
template<typename T> using min_priority_queue = priority_queue<T, vector<T>, greater<T>>;
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}
// 二項係数計算
long long COM(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}
ll gcd(ll x, ll y) {
   if (y == 0) return x;
   else if (y > x) {
       return gcd (y, x); 
   }
   else return gcd(x % y, y);
}
ll lcm(ll x, ll y) {
   return x / gcd(x, y) * y;
}
ll my_sqrt(ll x) {
    ll m = 0;
    ll M = 3000000001;
    while (M - m > 1) {
        ll now = (M + m) / 2;
        if (now * now <= x) {
            m = now;
        }
        else {
            M = now;
        }
    }
    return m;
}
ll keta(ll n) {
    ll ret = 0;
    while (n) {
        n /= 10;
        ret++;
    }
    return ret;
}
ll ceil(ll n, ll m) {
    // n > 0, m > 0
    ll ret = n / m;
    if (n % m) ret++;
    return ret;
}
ll pow_ll(ll x, ll n) {
    if (n == 0) return 1;
    if (n % 2) {
        return pow_ll(x, n - 1) * x;
    }
    else {
        ll tmp = pow_ll(x, n / 2);
        return tmp * tmp;
    }
}
vector<ll> compress(vector<ll> v) {
    // [3 5 5 6 1 1 10 1] -> [1 2 2 3 0 0 4 0] 
    vector<ll> u = v;
    sort(u.begin(), u.end());
    u.erase(unique(u.begin(),u.end()),u.end());
    map<ll, ll> mp;
    for (int i = 0; i < u.size(); i++) {
        mp[u[i]] = i;
    }
    for (int i = 0; i < v.size(); i++) {
        v[i] = mp[v[i]];
    }
    return v;
}
vector<P> v[100001];
bool visited1[100001];
bool visited2[100001];
P dfs1(int cur, int par, int type, ll num, ll k) {
    // type1   x + num         type2 num - x
    visited1[cur] = true;
    ll m, M;
    if (type == 1) {
        m = 1 - num;
        M = k - num;
    }
    else {
        m = num - k;
        M = num - 1;
    }
    for (auto x : v[cur]) {
        int nex = x.first;
        ll c = x.second;
        if (visited1[nex]) continue;
        P p = dfs1(nex, cur, 3 - type, c - num, k);
        m = max(p.first, m);
        M = min(p.second, M);
    }
    return P(m, M);
}
P dfs2(int cur, int par, int type, ll num, ll k) {
    // type1   x + num         type2 num - x
    visited2[cur] = true;
    ll m, M;
    if (type == 1) {
        m = 1 - num;
        M = k - num;
    }
    else {
        m = num - k;
        M = num - 1;
    }
    for (auto x : v[cur]) {
        int nex = x.first;
        ll c = x.second;
        if (visited2[nex]) continue;
        P p = dfs2(nex, cur, 3 - type, c - num, k);
        m = max(p.first, m);
        M = min(p.second, M);
    }
    return P(m, M);
}
int main() {
    for (int i = 0; i <= 100000; i++) {
        visited1[i] = false;
        visited2[i] = false;
    }
    ll n, m, k;
    cin >> n >> m >> k;
    dsu d(n);
    for (int i = 1; i <= m; i++) {
        ll x, y, z;
        cin >> x >> y >> z;
        x--;
        y--;
        v[x].push_back({y, z});
        v[y].push_back({x, z});
        d.merge(x, y);
    }
    mint ans = 1;
    mint ans2 = 1;
    for (auto g : d.groups()) {
        P p1 = dfs1(g[0], -1, 1, 0, k);
        mint tmp1 = max(p1.second - p1.first + 1, (ll)0);
        P p2 = dfs2(g[0], -1, 1, 0, k - 1);
        mint tmp2 = max(p2.second - p2.first + 1, (ll)0);
        // cout << g[0] << ' ' << tmp1.val() << ' ' << tmp2.val() << endl;
        ans *= tmp1;
        ans2 *= tmp2;
    }
    mint ret = ans - ans2;
    cout << ret.val() << endl;
    return 0;
}
הההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההה
XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
0