이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
using namespace std;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef long long ll;
typedef pair<ll,ll> pll;
typedef tuple<int,int,int> ti;
typedef unsigned long long ull;
typedef long double ld;
typedef vector<ll> vll;
typedef pair<ld,ld> pld;
#define pb push_back
#define popb pop_back()
#define pf push_front
#define popf pop_front
#define ff first
#define ss second
#define MOD (int)(1e8)
#define INF (ll) (1e18)
#define all(v) (v).begin(),(v).end()
ll gcd(ll a , ll b) {return b ? gcd(b , a % b) : a ;}
ll lcm(ll a , ll b) {return (a * b) / gcd(a , b);}
ld pointdist(ld x, ld y, ld point) { return ((x-point)*(y-point))/abs(x-y); }
//ld dist(ld x, ld y, ld a, ld b){ return sqrt((x-a)*(x-a) + (y-b)*(y-b)); }
const int nx[8] = {0, 0, 1, -1,1,1,-1,-1}, ny[8] = {1, -1, 0, 0,1,-1,1,-1}; //East, West, South, North+
////////////******SOLUTION******\\\\\\\\\\\
const int MAX_N = 1e5 + 4;
const int MAX_M = 2e5 + 4;
pll edges[MAX_N];
vector<pair<pll,int>> adj[MAX_N];
int n, m;
bool visited[MAX_N];
ll dist[MAX_N];
void dfs(int x)
{
visited[x] = true;
map<int,int> m;
map<ll,ll> sm;
for(auto e: adj[x])
{
int col = edges[e.ff.ss].ff;
m[col] ++;
sm[col] += edges[e.ff.ss].ss;
if(visited[e.ff.ff])
continue;
dfs(e.ff.ff);
}
for(auto e: adj[x])
{
int col = edges[e.ff.ss].ff;
if(m[col] == 1)
e.ss = 1;
}
}
ll dijkstra(int x)
{
set<pll> pq;
pq.insert({0,x});
while(!pq.empty())
{
pll pr = *(pq.begin());
ll node = pr.ss;
ll d = pr.ff;
pq.erase(pq.begin());
dist[node] = d;
visited[node] = true;
if(node == n)
return d;
for(auto e: adj[node])
{
if(visited[e.ff.ff])
continue;
if(e.ss)
{
if(dist[e.ff.ff] > d)
{
dist[e.ff.ff] = d;
pq.insert({d,e.ff.ff});
}
}
else
{
if(dist[e.ff.ff] > d+edges[e.ff.ss].ss)
{
dist[e.ff.ff] = d+edges[e.ff.ss].ss;
pq.insert({d+edges[e.ff.ss].ss,e.ff.ff});
}
}
}
}
return 69;
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
cin >> n >> m;
for(int i = 0; i <m; i ++)
{
ll a, b, c, p;
cin >> a >> b >> c >> p;
adj[a].pb({{b,i},0});
adj[b].pb({{a,i},0});
edges[i] = {c,p};
}
dfs(1);
if(!visited[n])
{
cout << -1;
exit(0);
}
memset(visited,false,sizeof(visited));
for(int i = 2; i <=n; i ++)
dist[i] = INF;
dist[1] = 0;
cout << dijkstra(1);
}
/*
Identify problem diagram: Brute force, Greedy, Dynamic Programming, Divide and Conquer
Reformulate the problem into something more theoretical
!!!!! IMPLICIT GRAPH ??????
!!!!! PAY ATTENTION TO THE CONSTRAINTS: DP nD ? BF ? BITMASKING ?
!!!!! SOLVE THE SUBTASKS FIRST: IT'S TOTALLY OK NOT TO SOLVE THE PROBLEM ENTIRELY
Search for multiple approaches: select the seemingly optimal one
Remember that you're the king of CP
Change your approach
Imagine Corner cases before submitting
Don't spend too much time on the problem: move on !
*/
컴파일 시 표준 에러 (stderr) 메시지
Main.cpp:26:1: warning: multi-line comment [-Wcomment]
26 | ////////////******SOLUTION******\\\\\\\\\\\
| ^
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