제출 #1233365

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1233365		clemmy14	자매 도시 (APIO20_swap)	C++20	0 / 100	0 ms	324 KiB

swap

#include "swap.h"
#include<bits/stdc++.h>
using namespace std;

int n;
// weight then node
vector<vector<pair<int, int>>> adj;

void init(int N, int M, vector<int> U, vector<int> V, vector<int> W) {
	n=N;
	adj = vector<vector<pair<int, int>>>(N);
	for(int i=0; i<M; i++) {
		adj[U[i]].push_back({W[i], V[i]});
		adj[V[i]].push_back({W[i], U[i]});
	}
	for(int i=0; i<N; i++) sort(adj[i].begin(), adj[i].end());
	
}

int ans=0;
set<pair<int, int>> noUse;

vector<int> dijkstra(int s, int e) {
	vector<int> par(n), dis(n, 1e9); dis[s]=0;
	vector<bool> vis(n, false);
	priority_queue<pair<int, int>> q; q.push({0, s});
	while(!q.empty()) {
		int node = q.top().second; q.pop();
		if(vis[node]) continue;
		vis[node]=true;
		if(node == e) {
			ans=dis[e]; break;
		}
		for(auto [w, child] : adj[node]) if(!noUse.count({node, child})) {
			if(dis[child] > max(dis[node], w)) {
				dis[child]=max(dis[node], w);
				par[child]=node;
				q.push({-dis[child], child});
			}
		}
	}
	vector<int> path; path.push_back(e);
	for(int i=e; i!=s; i=par[i]) {
		path.push_back(par[i]);
	}
	reverse(path.begin(), path.end());
	return path;
}

int getMinimumFuelCapacity(int X, int Y) {
	ans=0;	
	vector<int> path=dijkstra(X, Y);
	// for(auto x : path) cout << x << ' ';
	// cout << endl;
	int newAns=1e9;
	for(int i=0; i<path.size(); i++) {
		if(path[i] == X || path[i] == Y) {
			if(adj[path[i]].size() > 2) {
				if(adj[path[i]][2].first < ans) newAns=min(newAns, adj[path[i]][1].first);
				else newAns=min(newAns, adj[path[i]][2].first);
			}
		} else {
			bool f=false;
			for(int j=0; j<adj[path[i]].size(); j++) {
				if(!f && adj[path[i]][j].first == ans) f=true;
				else {
					newAns=min(newAns, adj[path[i]][j].first);
					break;
				}
			}
		}
	}

	//noUse.clear();
	for(int i=1; i<path.size(); i++) {
		noUse.insert({path[i], path[i-1]});
		noUse.insert({path[i-1], path[i]});
	}
	int oldAns=ans;
	ans=0;
	dijkstra(X, Y); noUse.clear();
	newAns=min(newAns, ans);

	return max(newAns, oldAns);
}

• Subtask 1: 0 / 6
• Subtask 2: 0 / 7
• Subtask 3: 0 / 17
• Subtask 4: 0 / 20
• Subtask 5: 0 / 23
• Subtask 6: 0 / 27