Problem
Concept & Idea
dfs를 이용해서 치킨집을 갯수만큼 정해주고, 만약 m과 치킨집의 수가 맞으면 각 좌표값을 이용하여 집집마다의 치킨 거리를 계산할 수 있는 문제였다.
Code
#include <iostream>
#include <queue>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;
int n,m;
vector<pair<int, int>> chicken, house;
long mins=9999999999;
void dfs(int x, vector<pair<int, int>> v){
if(v.size()==m) {
int nu=0;
for(int i=0; i<house.size(); i++) {
int distance = 101;
for(int j=0; j<v.size(); j++) {
if(abs(house[i].first-v[j].first)+abs(house[i].second-v[j].second)<distance)
distance=abs(house[i].first-v[j].first)+abs(house[i].second-v[j].second);
}
nu+=distance;
}
if(nu<mins)
mins=nu;
} else {
for(int j=x+1; j<chicken.size(); j++) {
v.push_back(chicken[j]);
dfs(j, v);
v.pop_back();
}
}
}
int main() {
cin>>n>>m;
for(int i=1; i<=n; i++) {
for(int j=1; j<=n; j++) {
int t;
cin>>t;
if(t==2) {
chicken.push_back({i,j});
}
if(t==1) {
house.push_back({i,j});
}
}
}
vector<pair<int, int>> ve;
ve.push_back(chicken[0]);
dfs(0,ve);
ve.pop_back();
dfs(0,ve);
cout<<mins<<endl;
}
Fealing
후하하하하핳
Check out this code in Victoria’s Gist. Please Comment my code in this link.