# | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
---|---|---|---|---|---|---|---|
390315 | oleksg | Robot (JOI21_ho_t4) | C++14 | 0 ms | 0 KiB |
이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#pragma GCC optimize("O2")
#include <fstream>
#include <string>
#include <iostream>
#include <bitset>
#include <math.h>
#include <string>
#include <algorithm>
#include <assert.h>
#include <bits/stdc++.h>
#include <vector>
#include <queue>
#include<stdio.h>
#include<ctype.h>
#define ll long long
using namespace std;
ll n, m;
struct con{
ll d;
ll color;
ll cost;
};
vector<con> cons[100001];
unordered_map<ll, ll> cost[100001];
// how much to convert all nodes of a color to some other color
unordered_map<ll, ll> dist[100001];
unordered_map<ll, ll> used[100001];
//if node with color x was used
int main(){
cin >> n >> m;
ll one, two, three, four;
for (int x = 0; x < m; x++){
cin >> one >> two >> three >> four;
cons[one - 1].push_back({two - 1, three, four});
cons[two - 1].push_back({one - 1, three, four});
}
for (int x = 0; x < n; x++){
for (int y = 0; y < cons[x].size(); y++){
ll col = cons[x][y].color;
ll cst = cons[x][y].cost;
if (cost[x].find(col) == cost[x].end()){
cost[x][col] = cst;
} else {
cost[x][col] += cst;
}
}
}
priority_queue<tuple<ll, ll, ll>> q;
//cost, node, color
q.push(make_tuple(0, 0, 0));
while(q.size() > 0){
ll node;
ll cst;
ll color;
tie(cst, node, color) = q.top();
q.pop();
if (used[node].find(color) == used[node].end() && cost == dist[node][color]){
used[node][color] = 1;
cst *= -1;
if (color != 0){
//we need to go to a node with the same color
for (auto con: cons[node]){
if (con.color == color){
if (used[con.d].find(0) == used[con.d].end()){
ll ncost = cst + cost[node][color] - con.cost;
if (dist[con.d].find(0) == dist[con.d].end() || dist[con.d][0] > ncost){
dist[con.d][0] = ncost;
q.push(make_tuple(-1 * ncost, con.d, 0));
}
}
}
}
}
else{
//there are 3 things which we can do
for (auto con: cons[node]){
if (used[con.d].find(con.color) == used[con.d].end()){
//dont switch anything and make the next node switch
ll ncost = cst;
if (dist[con.d].find(con.color) == dist[con.d].end() || dist[con.d][con.color] > ncost && dist[con.d][con.color] < dist[con.d][0] && cost[con.d][con.color] > con.cost){
dist[con.d][con.color] = ncost;
q.push(make_tuple(-1 * ncost, con.d, con.color));
}
}
if (used[con.d].find(0) == used[con.d].end()){
//switch everything but the connection
ll ncost = cst + cost[node][con.color] - con.cost;
if (dist[con.d].find(0) == dist[con.d].end() || dist[con.d][0] > ncost){
dist[con.d][0] = ncost;
q.push(make_tuple(-1 * ncost, con.d, 0));
}
//switch our node only
ncost = cst + con.cost;
if (dist[con.d].find(0) == dist[con.d].end() || dist[con.d][0] > ncost){
dist[con.d][0] = ncost;
q.push(make_tuple(-1 * ncost, con.d, 0));
}
}
}
}
}
}
if (dist[n - 1].find(0) == dist[n - 1].end()){
cout << -1 << "\n";
}
else{
cout << dist[n - 1][0] << "\n";
}
}