이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
/*
Code by @marlov
*/
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pi;
#define INF 200000000
#define maxV 30005
int N,M;
int dogea[maxV], dogeb[maxV];
//distance to location 0-N-1 inclusive
int dist[maxV];
bool vdoge[maxV];
priority_queue< pair<int,int> > pq;
//location contains array
vector<int> lc[maxV];
int main() {
ios_base::sync_with_stdio(0); cin.tie(0);
cin>>N>>M;
fill(dist,dist+N,INF);
for(int i=0;i<M;i++){
cin>>dogea[i]>>dogeb[i];
lc[dogea[i]].push_back(i);
}
/*for(int j=0;j<lc[dogea[0]].size();j++){
pq.push(make_pair(0,lc[dogea[0]][j]));
}*/
pq.push(make_pair(0, dogea[0]));
dist[dogea[0]]=0;
while(!pq.empty()){
//cd is number of jumps and ci is the current index
int cd=-pq.top().first;
int si=pq.top().second;
pq.pop();
//if that dog has already been visited there is no need to check again
if(vdoge[si]) continue;
vdoge[si]=true;
for(int j=0;j<lc[si].size();j++){
int ci = lc[si][j];
//loops from current location to 0
for(int i=0, cl=dogea[ci];cl>=0;i++, cl -= dogeb[ci]){
//current location
if(cd+i<dist[cl]){
dist[cl]=cd+i;
/*for(int j=0;j<lc[cl].size();j++){
pq.push(make_pair(-dist[cl],lc[cl][j]));
}*/
pq.push(make_pair(-dist[cl],cl));
}
}
//to N-1
for(int i=1;dogea[ci]+i*dogeb[ci]<N;i++){
int cl=dogea[ci]+i*dogeb[ci];
if(cd+i<dist[cl]){
dist[cl]=cd+i;
/*for(int j=0;j<lc[cl].size();j++){
pq.push(make_pair(-dist[cl],lc[cl][j]));
}*/
pq.push(make_pair(-dist[cl],cl));
}
}
}
}
if(dist[dogea[1]]==INF){
cout<<-1<<'\n';
}else{
cout<<dist[dogea[1]]<<'\n';
}
return 0;
}
/* stuff you should look for
* int overflow, array bounds
* special cases (n=1,n=0?)
* do smth instead of nothing and stay organized
*/
컴파일 시 표준 에러 (stderr) 메시지
skyscraper.cpp: In function 'int main()':
skyscraper.cpp:45:22: warning: comparison between signed and unsigned integer expressions [-Wsign-compare]
for(int j=0;j<lc[si].size();j++){
~^~~~~~~~~~~~~~
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