SOL)
BFS 어떤 순열과 이 순열에서 만들 수 있는 새로운 순열사이 간선이 연결된
Graph라 생각하고 문제를 푼다.
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#include<cstdio>
#include<vector>
#include<queue>
#include<algorithm>
#include<map>
using namespace std;
map<vector<int>, int> visit;
vector<int> arr, goal;
int N, K;
int BFS(){
int cost = 0;
queue<vector<int> > q;
q.push(arr);
visit[arr]++;
while (!q.empty()){
int size = q.size();
for (int l = 0; l < size; l++){
vector<int> mv = q.front();
q.pop();
if (mv == goal) return cost;
for (int i = 0; i <= N - K; i++){
reverse(mv.begin() + i, mv.begin() + i + K);
if (visit.count(mv) == 0){
visit[mv]++;
q.push(mv);
}
reverse(mv.begin() + i, mv.begin() + i + K);
}
}
cost++;
}
return -1;
}
int main(){
scanf("%d%d", &N, &K);
arr = vector<int>(N);
for (int i = 0; i < N; i++)
scanf("%d", &arr[i]);
goal = arr;
sort(goal.begin(), goal.end());
printf("%d", BFS());
}
#include<vector>
#include<queue>
#include<algorithm>
#include<map>
using namespace std;
map<vector<int>, int> visit;
vector<int> arr, goal;
int N, K;
int BFS(){
int cost = 0;
queue<vector<int> > q;
q.push(arr);
visit[arr]++;
while (!q.empty()){
int size = q.size();
for (int l = 0; l < size; l++){
vector<int> mv = q.front();
q.pop();
if (mv == goal) return cost;
for (int i = 0; i <= N - K; i++){
reverse(mv.begin() + i, mv.begin() + i + K);
if (visit.count(mv) == 0){
visit[mv]++;
q.push(mv);
}
reverse(mv.begin() + i, mv.begin() + i + K);
}
}
cost++;
}
return -1;
}
int main(){
scanf("%d%d", &N, &K);
arr = vector<int>(N);
for (int i = 0; i < N; i++)
scanf("%d", &arr[i]);
goal = arr;
sort(goal.begin(), goal.end());
printf("%d", BFS());
}
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