# | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
---|---|---|---|---|---|---|---|
263546 | MKopchev | Olympic Bus (JOI20_ho_t4) | C++14 | 758 ms | 4048 KiB |
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;
const int nmax=200+5,mmax=50000+5;
const long long inf=1e16+42;
int n,m;
struct edges
{
int from,to,cost,to_rev;
};
edges inp[mmax];
vector< pair<int/*dist*/,int/*to*/> > adj[nmax];
long long dist[nmax];
priority_queue< pair<long long/*dist*/,int/*node*/> > pq,idle;
long long dij(int from,int to)
{
memset(dist,-1,sizeof(dist));
pq=idle;
pq.push({0,from});
while(dist[to]==-1&&pq.size())
{
pair<long long/*dist*/,int/*node*/> tp=pq.top();
pq.pop();
tp.first=-tp.first;
if(dist[tp.second]!=-1)continue;
dist[tp.second]=tp.first;
for(auto k:adj[tp.second])
if(dist[k.second]==-1)pq.push({-(tp.first+k.first),k.second});
}
//for(int i=1;i<=n;i++)cout<<dist[i]<<" ";cout<<endl;
if(dist[to]==-1)return inf;
return dist[to];
}
long long solve(int which)
{
long long ret=inp[which].to_rev;
for(int i=1;i<=n;i++)adj[i]={};
for(int i=1;i<=m;i++)
if(i!=which)adj[inp[i].from].push_back({inp[i].cost,inp[i].to});
else adj[inp[i].to].push_back({inp[i].cost,inp[i].from});
ret=ret+dij(1,n);
ret=ret+dij(n,1);
//cout<<"which= "<<which<<" ret= "<<ret<<endl;
return ret;
}
long long floyd[nmax][nmax];
int main()
{
scanf("%i%i",&n,&m);
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
floyd[i][j]=inf;
for(int i=1;i<=n;i++)
floyd[i][i]=0;
for(int i=1;i<=m;i++)
{
scanf("%i%i%i%i",&inp[i].from,&inp[i].to,&inp[i].cost,&inp[i].to_rev);
floyd[inp[i].from][inp[i].to]=min(floyd[inp[i].from][inp[i].to],1LL*inp[i].cost);
}
for(int k=1;k<=n;k++)
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
floyd[i][j]=min(floyd[i][j],floyd[i][k]+floyd[k][j]);
/*
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)cout<<i<<" "<<j<<" -> "<<floyd[i][j]<<endl;
*/
long long outp=inf;
outp=min(outp,floyd[1][n]+floyd[n][1]);
for(int i=1;i<=m;i++)
{
long long low_path_1=min(floyd[1][n],floyd[1][inp[i].to]+inp[i].cost+floyd[inp[i].from][n]);
long long low_path_2=min(floyd[n][1],floyd[n][inp[i].to]+inp[i].cost+floyd[inp[i].from][1]);
//cout<<i<<" -> "<<low_path_1<<" "<<low_path_2<<" "<<inp[i].to_rev<<" "<<outp<<endl;
if(low_path_1+low_path_2+inp[i].to_rev<outp)outp=min(outp,solve(i));
}
if(outp>=inf)printf("-1\n");
else printf("%i\n",outp);
return 0;
}
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