結果
| 問題 |
No.1364 [Renaming] Road to Cherry from Zelkova
|
| コンテスト | |
| ユーザー |
|
| 提出日時 | 2021-01-22 23:27:05 |
| 言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
| 結果 |
AC
|
| 実行時間 | 200 ms / 2,500 ms |
| コード長 | 4,511 bytes |
| コンパイル時間 | 2,436 ms |
| コンパイル使用メモリ | 151,528 KB |
| 最終ジャッジ日時 | 2025-01-18 06:10:48 |
|
ジャッジサーバーID (参考情報) |
judge3 / judge3 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 3 |
| other | AC * 45 |
ソースコード
#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
#include <queue>
#include <string>
#include <map>
#include <set>
#include <stack>
#include <tuple>
#include <deque>
#include <array>
#include <numeric>
#include <bitset>
#include <iomanip>
#include <cassert>
#include <chrono>
#include <random>
#include <limits>
#include <iterator>
#include <functional>
#include <sstream>
#include <fstream>
#include <complex>
#include <cstring>
#include <unordered_map>
#include <unordered_set>
using namespace std;
using ll = long long;
constexpr int INF = 1001001001;
constexpr int mod = 1000000007;
// constexpr int mod = 998244353;
template<class T>
inline bool chmax(T& x, T y){
if(x < y){
x = y;
return true;
}
return false;
}
template<class T>
inline bool chmin(T& x, T y){
if(x > y){
x = y;
return true;
}
return false;
}
struct mint {
int x;
mint() : x(0) {}
mint(int64_t y) : x(y >= 0 ? y % mod : (mod - (-y) % mod) % mod) {}
mint& operator+=(const mint& p){
if((x += p.x) >= mod) x -= mod;
return *this;
}
mint& operator-=(const mint& p){
if((x -= p.x) < 0) x += mod;
return *this;
}
mint& operator*=(const mint& p){
x = (int)(1LL * x * p.x % mod);
return *this;
}
mint& operator/=(const mint& p){
*this *= p.inverse();
return *this;
}
mint operator-() const { return mint(-x); }
mint operator+(const mint& p) const { return mint(*this) += p; }
mint operator-(const mint& p) const { return mint(*this) -= p; }
mint operator*(const mint& p) const { return mint(*this) *= p; }
mint operator/(const mint& p) const { return mint(*this) /= p; }
bool operator==(const mint& p) const { return x == p.x; }
bool operator!=(const mint& p) const { return x != p.x; }
mint pow(int64_t n) const {
mint res = 1, mul = x;
while(n > 0){
if(n & 1) res *= mul;
mul *= mul;
n >>= 1;
}
return res;
}
mint inverse() const { return pow(mod - 2); }
friend ostream& operator<<(ostream& os, const mint& p){
return os << p.x;
}
friend istream& operator>>(istream& is, mint& p){
int64_t val;
is >> val;
p = mint(val);
return is;
}
};
int main(){
ios::sync_with_stdio(false);
cin.tie(nullptr);
int N, M;
cin >> N >> M;
using edge = tuple<int, int, int>;
vector<vector<edge>> g(N + 1), rg(N + 1);
for(int i = 0; i < M; ++i){
int u, v, l, a;
cin >> u >> v >> l >> a;
g[u].emplace_back(v, l, a);
rg[v].emplace_back(u, l, a);
}
vector<bool> from0(N + 1), fromN(N + 1), visited(N + 1);
auto dfs = [&](auto&& self, vector<vector<edge>>& G, vector<bool>& f, int cur) -> void {
visited[cur] = true;
f[cur] = true;
for(auto [nxt, cost, weight] : G[cur]){
if(!visited[nxt]) self(self, G, f, nxt);
}
};
dfs(dfs, g, from0, 0);
visited.assign(N + 1, false);
dfs(dfs, rg, fromN, N);
int V = 0;
vector<bool> flag(N + 1);
for(int i = 0; i <= N; ++i){
flag[i] = from0[i] && fromN[i];
V += flag[i];
}
vector<int> indeg(N + 1);
for(int from = 0; from <= N; ++from){
if(!flag[from]) continue;
for(auto [to, cost, weight] : g[from]){
if(flag[to]) indeg[to] += 1;
}
}
queue<int> que;
for(int i = 0; i <= N; ++i){
if(flag[i] && indeg[i] == 0) que.emplace(i);
}
vector<int> topological_order;
while(!que.empty()){
int from = que.front();
que.pop();
topological_order.emplace_back(from);
for(auto [to, cost, weight] : g[from]){
if(!flag[to]) continue;
if(--indeg[to] == 0) que.emplace(to);
}
}
if((int)topological_order.size() != V){
cout << "INF\n";
return 0;
}
vector<mint> dp(N + 1), dp2(N + 1);
dp2[0] = 1;
for(int from : topological_order){
for(auto [to, cost, weight] : g[from]){
if(!flag[to]) continue;
dp2[to] += dp2[from] * weight;
}
}
for(int from : topological_order){
for(auto [to, cost, weight] : g[from]){
if(!flag[to]) continue;
dp[to] += (dp[from] + dp2[from] * cost) * weight;
}
}
cout << dp[N] << endl;
return 0;
}