結果

問題 No.2739 Time is money
ユーザー The Forsaking
提出日時 2025-01-22 16:56:05
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
RE  
実行時間 -
コード長 3,454 bytes
コンパイル時間 1,068 ms
コンパイル使用メモリ 123,120 KB
実行使用メモリ 29,392 KB
最終ジャッジ日時 2025-01-22 16:56:19
合計ジャッジ時間 12,155 ms
ジャッジサーバーID
（参考情報） 		judge4 / judge3
このコードへのチャレンジ
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ファイルパターン 結果
sample RE * 2
other RE * 18
権限があれば一括ダウンロードができます
コンパイルメッセージ
main.cpp: In function ‘int solve()’:
main.cpp:89:1: warning: no return statement in function returning non-void [-Wreturn-type]
   89 | }
      | ^
main.cpp: In function ‘int main()’:
main.cpp:97:14: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
   97 |         scanf("%d%d%d%d", &a, &b, &d, &c);
      |         ~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~~~~

ソースコード

diff #
プレゼンテーションモードにする

#include <iostream>
#include <sstream>
#include <iomanip>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <vector>
#include <queue> 
#include <unordered_set>
#include <unordered_map>
#include <bitset>
#include <ctime>
#include <assert.h>
#include <deque>
#include <list>
#include <stack>
using namespace std;
#define is_mul_overflow(a, b) \
    ((b != 0) && (a > LLONG_MAX / b || a < LLONG_MIN / b))
 
typedef pair<long long, int> pli;
typedef pair<int, long long> pil;
typedef pair<long long , long long> pll;
typedef pair<int, int> pii;
typedef pair<double, double> pdd;
typedef pair<int, pii> piii;
typedef pair<int, long long > pil;
typedef pair<long long, pii> plii;
typedef pair<double, int> pdi;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ull, ull> puu;
typedef long double ld;
const int N = 2000086, MOD = 998244353, INF = 0x3f3f3f3f, MID = 333;
const long double EPS = 1e-8;
int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
char p[10] = "ULDR";
// int dx[8] = {1, 1, 0, -1, -1, -1, 0, 1}, dy[8] = {0, 1, 1, 1, 0, -1, -1, -1};
// int dx[8] = {2, 1, -1, -2, -2, -1, 1, 2}, dy[8] = {1, 2, 2, 1, -1, -2, -2, -1};
int n, m, cnt;
int w[N];
vector<ll> num;
ll res;
ll lowbit(ll x) { return x & -x; }
ll lcm(ll a, ll b) { return a / __gcd(a, b) * b; }
inline double rand(double l, double r) { return (double)rand() / RAND_MAX * (r - l) + l; }
inline ll qmi(ll a, ll b, ll c) { ll res = 1; while (b) { if (b & 1) res = res * a % c; a = a * a % c; b >>= 1; } return res; }
inline ll qmi(ll a, ll b) { ll res = 1; while (b) { if (b & 1) res *= a; a *= a; b >>= 1; } return res; }
inline double qmi(double a, ll b) { double res = 1; while (b) { if (b & 1) res *= a; a *= a; b >>= 1; } return res; } 
// inline ll C(ll a, ll b) { if (a < b) return 0; if (b > a - b) b = a - b; ll res = 1; for (ll i = 1, j = a; i <= b; i++, j--) { res = res * (j % 
    MOD) % MOD; res = res * qmi(i, MOD - 2, MOD) % MOD; } return res; }
inline ll C(ll a, ll b, int* c) { if (a < b) return 0; ll res = 1; for (ll j = a, i = 1; i < b + 1; i++, j--) res *= j; for (ll j = a, i = 1; i < b + 
    1; i++, j--) res /= i; return res; }
inline int find_(int x) { return lower_bound(num.begin(), num.end(), x) - num.begin(); }
int e[N], ne[N], h[N], v[N], idx, x;
ll d[N], t[N];
bool st[N];
inline void add(int a, int b, int c, int d) { e[idx] = b, ne[idx] = h[a], w[idx] = c, v[idx] = d, h[a] = idx++; }
int solve() {
    d[1] = 0;
    priority_queue<plii, vector<plii>, greater<plii>> q;
    q.push({0, {0, 1}});
    while (q.size()) {
        auto it = q.top(); q.pop();
        int u = it.second.second;
        if (st[u]) continue;
        st[u] = 1;
        for (int i = h[u]; ~i; i = ne[i]) {
            int j = e[i];
            if (st[j]) continue;
            ll nd = d[u] + w[i], nt = t[u] + v[i];
            nd += nt / x, nt %= x;
            if (d[j] > nd || d[j] == nd && t[j] > nt) {
                d[j] = nd, t[j] = nt;
                q.push({d[j], {t[j], j}});
            }
        }
    }
    if (d[n] > 1e16) puts("-1");
    else printf("%lld\n", d[n] + !!t[n]);
}
int main() {
    cin >> n >> m >> x;
    memset(h, -1, sizeof(int) * (n + 10));
    memset(d, 0x3f, sizeof(ll) * (n + 10));
    for (int i = 1; i < m + 1; i++) {
        int a, b, c, d;
        scanf("%d%d%d%d", &a, &b, &d, &c);
        add(a, b, c, d), add(b, a, c, d);
    }
    solve();
    return 0;
}
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