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BOJ_algorithm
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13700.cpp
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#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
/*
입력
첫째 줄에 N, S, D, F, B, K가 주어지고,
K > 0인 경우에는 둘째 줄에 경찰서의 위치 l1, l2, …, lK가 주어진다.
(1 ≤ S, D, k ≤ N ≤ 100000, 0 ≤ F, B ≤ 100000, 0 ≤ K ≤ N/2, S ≠ D ≠ l)
출력
첫째 줄에 대도 X가 건물 S에서 자신의 집 D로 무사히 가기 위해
지언이가 버튼을 눌러야 하는 최소 횟수를 출력한다
. 만약, D에 도달할 수 없는 데이터인 경우 “BUG FOUND”를 출력한다.
*/
int d[100001];
bool v[100001];
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int N,S,D,F,B,K,x;
cin>>N>>S>>D>>F>>B>>K;
for(int i=0;i<K;i++)
{
cin>>x;
d[x]=-1;
v[x]=true;
}
queue<int> q;
q.push(S);
v[S]=true;
while(!q.empty())
{
int p=q.front();//current position
int fp=p+F;
int bp=p-B;
q.pop();
if(fp>=1&&fp<=N&&v[fp]==false)
{
v[fp]=true;
d[fp]=d[p]+1;
q.push(fp);
}
if(bp>=1&&bp<=N&&v[bp]==false)
{
v[bp]=true;
d[bp]=d[p]+1;
q.push(bp);
}
}
if(d[D]<=0)
cout<<"BUG FOUND";
else
cout<<d[D];
}
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