# | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
---|---|---|---|---|---|---|---|
1195001 | nikulid | 이상적인 도시 (IOI12_city) | C++20 | 0 ms | 0 KiB |
#include <climits>
/*
lemma: I will solve subtask 4
*/
ll ans=0;
vector<vector<int>> visitted, exists; // both are [y][x]
int bfs(int I, int x, int y, int curdist){
// coords is [x,y]
visitted[y][x] = I;
ans = ans%1000000000;
if(exists[y+1][x+1] && visitted[y+1][x+1] != I){
ans += curdist+1;
bfs(I, x+1, y+1, curdist+1);
}
if(exists[y+1][x-1] && visitted[y+1][x-1] != I){
ans += curdist+1;
bfs(I, x-1, y+1, curdist+1);
}
if(exists[y-1][x+1] && visitted[y-1][x+1] != I){
ans += curdist+1;
bfs(I, x+1, y-1, curdist+1);
}
if(exists[y-1][x-1] && visitted[y-1][x-1] != I){
ans += curdist+1;
bfs(I, x-1, y-1, curdist+1);
}
return;
}
int DistanceSum(int N, int *X, int *Y) {
visitted = vector<vector<int>>(N+1, N+1, 0);
exists = vector<vector<int>>(N+1, N+1, 0);
// find min X and Y values, to translate the graph to start at (1,1)
int minX=INT_MAX-1, minY=INT_MAX-1;
for(int i=0; i<N; i++){
minX = min(minX, X[i]);
minY = min(minY, Y[i]);
}
for(int i=0; i<N; i++){
exists[Y[i]-minY][X[i]-minX] = 1;
}
for(int i=0; i<N; i++){
bfs(i, X[i], Y[i], 0);
}
return ans;
}