結果
問題 | No.1995 CHIKA Road |
ユーザー |
|
提出日時 | 2022-07-10 08:52:17 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 256 ms / 2,000 ms |
コード長 | 4,329 bytes |
コンパイル時間 | 3,149 ms |
コンパイル使用メモリ | 218,932 KB |
最終ジャッジ日時 | 2025-01-30 06:07:14 |
ジャッジサーバーID (参考情報) |
judge4 / judge1 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 37 |
ソースコード
#include<bits/stdc++.h>#include <unordered_map>#define int long long#define ll long long#define rep(i, n) for(ll i = 0; i < (n); i++)#define rep2(i, n) for(ll i = 0; n; i++)#define rep3(i,s,n) for(ll i = (s); i < (n); i++)#define rep4(i,s,n) for(ll i = (s); n; i++)#define repe(v, V) for(auto& v: V)#define all(x) (x).begin(),(x).end()#define v vector#define mod(a,b) (((a) % (b) + (b)) % (b)) /* 負の数に対応した剰余 */#define pi 3.14159265358979#define eps 1e-13#define inf 9223372036854775807#define MAX_LL 9223372036854775807#define umap unordered_mapusing namespace std;const int dx[4] = { 1, 0, -1, 0 };const int dy[4] = { 0, 1, 0, -1 };struct Edge { int to; int cost = 1; };using Graph = vector<vector<Edge>>;template<typename T> using min_priority_queue = priority_queue<T, vector<T>, greater<T>>;int to_base10(string Nx, int base) /* N進数から10進数に変換 */ { int N10 = 0; rep(i, Nx.length()) N10 += stoll(Nx.substr(Nx.length() - 1 - i, 1)) *pow(base, i); return N10; }string to_baseN(int N10, int base, int min_digits = 1) /* 10進数からN進数に変換 */ { string r; while (N10 != 0) { r = to_string(N10 % base) + r; N10/= base; } while (r.length() < min_digits) r = "0" + r; return r; }v<pair<int, int>> prime_factorize(int N) /* 素因数分解 .first: 素数, .second: 指数 */ { v<pair<int, int>> res; rep4(a, 2, a * a <= N) { if (N % a !=0) continue; int ex = 0; while (N % a == 0) { ++ex; N /= a; } res.push_back({a, ex}); } if (N != 1) res.push_back({N, 1}); return res; }v<int> divisors(int N)/*約数列挙*/ {v<int> r;rep4(i, 1, i * i <= N) {if (N % i == 0) {r.push_back(i);if(i != N / i) r.push_back(N / i);}}return r;}int modinv(int a, int m)/*逆元*/ { int b = m, u = 1, v = 0;while (b) { long long t = a / b;a -= t * b; swap(a, b); u -= t * v; swap(u, v); }u %= m;if (u < 0) u += m; return u;}int sigma(int a, int b)/*a,a+1,...,bの総和*/ {return b * (b + 1) / 2 - a * (a - 1) / 2;}v<v<int>> Pascal(int N, int mod = MAX_LL)/*パスカルの三角形(nCk=v[n][k]) O(N^2)*/ { v<v<int>> com(N + 1,v<int>(N + 1)); com[0][0] = 1; for (int i = 1; i < N + 1; ++i) { com[i][0] = 1; for (int j = 1; j < N + 1; j++) { com[i][j] = (com[i - 1][j - 1] + com[i - 1][j]) % mod; } }return com; }int Pow(int n, int p, int mod = MAX_LL)/* 指数関数のLL版(繰り返し2乗法) modあり */ {int ans = 1;int now = n;for (; p > 0; p >>= 1){if ((p & 1) == 1){ans *= now;ans %= mod;}now *= now;now %= mod;}return ans;}void printVec(const vector<int>& vec)/* vectorの中身をprint */ {cout << "";for (auto it = vec.begin(); it != vec.end(); ++it) {cout << *it << " "; }cout << endl;}void makeset(v<int>& vec)/* vectorをソートし、重複要素を削除*/ { sort(vec.begin(), vec.end()); vec.erase(unique(vec.begin(), vec.end()), vec.end());}int N, M;vector<int> dijkstra(const Graph& G, int start) {using P = pair<int, int>;const int INF = numeric_limits<int>::max();vector<int> dist(G.size(), INF);int N = G.size();dist.resize(N, INF);priority_queue<P, vector<P>, greater<P>> pq; // 「仮の最短距離, 頂点」が小さい順に並ぶdist[start] = 0;pq.emplace(dist[start], start);while (!pq.empty()) {P p = pq.top();pq.pop();int v = p.second;if (dist[v] < p.first) { // 最短距離で無ければ無視continue;}for (auto& e : G[v]) {if (dist[e.to] > dist[v] + e.cost) { // 最短距離候補なら priority_queue に追加dist[e.to] = dist[v] + e.cost;pq.emplace(dist[e.to], e.to);}}}return dist;}signed main(){cout << setprecision(15);cin >> N >> M;v<int> A(M), B(M), NodeList(0);rep(i, M) {cin >> A[i] >> B[i];NodeList.push_back(A[i]);NodeList.push_back(B[i]);}NodeList.push_back(1);NodeList.push_back(N);makeset(NodeList);int N2 = NodeList.size();Graph G(N2);rep(i, N2 - 1) {G[i].push_back({i + 1, (NodeList[i + 1] - NodeList[i]) * 2 });}rep(i, M) {int a = lower_bound(all(NodeList), A[i]) - NodeList.begin();int b = lower_bound(all(NodeList), B[i]) - NodeList.begin();G[a].push_back({ b, 2 * (B[i] - A[i]) - 1 });}auto ans = dijkstra(G, 0);cout << ans[N2 - 1];}